Chapter-2, POLYNOMIALS

Explanation: 

The correct response is C. (a) Because the exponent of x is - 2, which is not a whole number, the equation x222x2=x222x2 is not a polynomial.

(b) √2x−1=√2x12−1, Because the exponent of x is 12, which is not a whole integer, it is not a sine wave.

(c) Because each of the exponents of x are whole numbers, x2+3 x 32x = x2+3x 3212 =x2+3x is a polynomial.

(d) x−1x+1 Because it is a rational function, it is not a polynomial.

Explanation: 

It follows that √2

It can be expressed as √2 x 0.

The polynomial's degree in this case is 0.

Consequently, the polynomial in question has a degree of 0.

Do This: The polynomial with a value of 3 has degrees of a. 2, b. 0, c. 1, and d.

It is known that three.

It can be expressed as 3x0.

The polynomial's degree in this case is 0.

Consequently, the polynomial that is presented has a degree of 0.

Explanation: 

The right answer is option 4: 

4x4+0x3+0x5+5x+7=4x4+5x+7.

The maximum power of variable x in the non-zero term is similar to the degree of the expression 4x4+0x3+0x5+5x+7.

In this case, the expression 4x4's maximum power of x is 4.

Consequently, a polynomial has a degree of 4.

Explanation: 

We are aware that a polynomial's degree is defined as the highest sum of its exponents.

A term's degree, which is a non-negative integer, is calculated by adding the exponents of all the variables that appear in it.

Zero polynomials are any polynomials that have all of their variables' coefficients set to zero.

As an illustration, consider the polynomial 3x4 + 0x3 + 0x5 + 6x + 5.

The zero polynomial's degree is not specified.

As a result, the zero polynomial's degree is not known.

Explanation: 

The formula p(x) = x ² - 2x2 + 1 is known.

We must determine p(22).

Let's replace x by writing it as 22 p(2√2) = (2√2).² - 2√2(2√2) + 1

Further computation yields p(22) = 8 - 8 + 1.

So, we have p(22) = 1.

Consequently, p(22) equals 1.

Explanation: 

The formula p(x) = 5x - four times two + 3 is known.

Finding the value for p(-1) = 5(-1) - 4(-1)2 + 3 when x = -1 is necessary.

P(-1) is further calculated as -5 - 4 + 3

Therefore, we obtain p(-1) = -9 + 3 p(-1) = -6.

Consequently, the result is -6.

Explanation: 

Provided it means p(x) = x + 3.

Finding the ratio of p(x) + p(-x) is necessary.

In this section p(-x) = (-x + 3) and p(x) + p(-x) = (x + 3)

Consequently, we have p(x) + p(-x) = x + 3 - x + 3 and p(x) + p(-x) = 6.

Consequently, p(x) + p(-x) = 6.

Explanation: 

The ideal selection is C.

any genuine address

the justification behind the ideal choice.

Any cubic whose coefficients are all equal to zero is said to be a zero polynomial.

The value of a variable at which a polynomial becomes zero is known as the polynomial's zero.

The symbol for a zero polynomial is f(x)=0.

Each real integer is a zero of the zero polynomial as a result.

As a result, choice (c) is accurate.

Explanation: 

The response is B.

P(x) = 2x + 5 provided

Put p(x)=0 to identify the polynomial's zero. 2x+5=0 ⇒ x=−52

Therefore, 52 is the polynomial's value of zero.

Explanation: 

Getting the solution to the problem 2x2+7x4=0 is the same as solving 2x2+7x4=0.

⇒2x2+8x−x−4=0

⇒2x(x+4)−1(x+4)=0

 ⇒(2x−1)(x+4)=0

 ⇒(2x−1)=0,(x+4)=0

 ⇒x=21,−4

As a result, choice B is accurate.

Explanation: 

The solution is D.

Let p(x) = x51 + x51.(i) The residue obtained by dividing p(x) by x+1 is the same as p(-1).

With x=-1 in equation (i), we have p(1)=151+51=1+51=50.

Thus, the remaining amount is 50.

Explanation: 

Part p(x) is (x + 1), where x+1=0 and x=1.

p(1) = 0 2(1)2 + k(1) = 0 2 k = 0 k = 2 as a result.

Consequently, k has a value of 2.

Explanation: 

The response is B. x3+x2+x+1

Principle for variables P(a)=0 if (x-a) is a factor of P(x).

Let's examine each possibility to see if (x+1) is a factor of the provided polynomial.

Option A: Assume that the factor (x+1) is equal to x3+x2x+1.

So, x=1 is equal to zero when x3+x2x+1.

⇒(−1)3+(−1)2−(−1)+1 =−1+1+1=−1≠0

Therefore, Option A is untrue.

Option B: Assume that the factor (x+1) equals x3+x2+x+1.

As a result, x=1 is equal to zero when compared to x3+x2+x+1.

⇒ (−1)3+(−1)2+(−1)+1 

=−1+1−1+1=0

Option B is thus correct.

Assume that (x+1) is a factor of x4+x3+x2+1 in Option C.

As a result, x=1 is equal to zero when multiplied by x4+x3+x2+1.

⇒ (−1)4+(−1)3+(−1)2+1 

=1−1+1+1=2≠0

Option C is therefore untrue.

Assume that (x+1) is a factor of x4+3x3+3x2+x+1 in Option D.

Since x=1 is equal to zero, x4+3x3+3x2+x+1

⇒ (−1)4+3(−1)3+3(−1)2+(−1)+1

 =1−3+3−1+1=1≠0

Option D is therefore erroneous.

Explanation: 

The solution is D.

(25x2–1)+(1+5x)2

 =25x2−1+1+25x2+10x Validity states that (a+b)2=a2+b2+2ab.

=50x2+10x=10x(5x+1)

As a result, 10x is one of the factors of the provided polynomial.

Explanation: 

It follows that 2492 - 2482

Equating (249 + 248) (249 - 248) using the algebraic identity a2 - b2 = (a + b) (a - b)

Additional computation yields: = 497 x 1 = 497

The score is 497 as a result.

Explanation: 

We know that 4x2 + 8x + 3

divided as follows: = 4x2 + 6x + 2x + 3

Leaving out typical phrases, the formula is: 2x(2x + 3) + 1 (2x + 3)

The result is = (2x + 3) (2x + 1).

The solution is (2x + 3) (2x + 1) as a result.

Explanation: 

The right answer is C 3xy.

Allow P(x) to be (x+y)(x+y)2 (x+y)(x2xy+y2)

=(x+y)(x2+2xy+y2)−(x+y)(x2−xy+y2)

=(x+y)(x2+2xy+y2−x2+xy−y2)

[Using (x+y) as the default]

=(x+y)(3xy)

As a result, 3xy is a factor of the specified polynomial from the possibilities supplied.

Explanation: 

Using the formula (a+b)3=a3+b3+3ab(a+b), the solution (x+3)3=x3333+3x(3)(x+3)

=x3+27+9x(x+3) =x3+27+9x2+27x

As a result, 27 is the ratio of x in (x+3)3.

The solution is D, thus.

Explanation: 

The right response is C: 0.

We are aware that x3y3=(xy)(x2+xy+y2)...(i)

As a result, xy+yx=1

⇒x2+y2xy=−1

⇒x²+y²=−xy

When we add xy on both sides, we have x2+22+xy=0. When eq.(ii) is substituted for eq.(i), the result is x3y3=(xy)(0) x3y3=0.

Explanation: 

The equation 49x² - b = (7x + 1/2) is known. (7x - 1/2)

With the help of the mathematical formula a² - b² = (a + b) (a - b),

49x² - b = 49x² - 1/4

Using the equation on each side, b = ¼.

Explanation: 

We are aware that a3 + b3 - c3 = (a + b + c) (a², b², and c²) - bc, ca, and ab

The formula a + b + c = 0 is known.

As a result, we have a3 + b3 + c3 - 3abc = 0 and a3 + b3 + c3 = 3abc.

As a result, 3abc is equivalent to a3, b3, and c3.

Explanation: 

A polynomial may contain exponents, constants, and variables, but it never divides by a variable.

There cannot be a negative exponent in a polynomial.

No variable at all, one variable, two variables, or more are all possible for polynomials. Exponents can only be multiples of 0, 1, or 2, not fractions.

According to the definition given above, (iii) 1-5x is not a polynomial as its exponent is a fraction.

Due to the negative exponent, (v) x(x-2)(x-4)/x , (vi) 1/1+x , and (viii) 1/2x  are not polynomials.

The polynomials are therefore (i), (ii), (iv), and (vii).

Explanation: 

I False.

A binomial has two components, exactly.

ii False.

Not all polynomials are binomials.

Using the example of (a) 3x²+4x+5 [polynomial but not a binomial](b) 3x²+5, which is polynomial and binomial.

True. iii.

A polynomial with a degree of a whole number larger than one is called a binomial. Thus, it might be of degree 5.

iv. Untrue.

Any real number can serve as the polynomial's zero.

True, False

There can be any number of zeros in a polynomial. It depends on the polynomial's degree.

v. Untrue.

The sum of any two polynomials with the same degree does not always have the same degree.

Let f(x)=x4+2 and g(x)=x4+4x3+2x, for instance.

Two polynomials added together, f(x)+g(x)=x4+2+(x4+4x3+2x)=4x3+2x+2 which is not a degree four polynomial.

Explanation: 

The polynomial is xy + yz + zx, given.

The provided polynomial has to be identified.

A polynomial is an expression that solely uses the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. It is made up of variables and coefficients.

Based on how many variables they contain, polynomials can be categorized.

1) A monomial statement has a single variable.

2) A two-variable expression known as a binomial

A three-variable expression is referred to as a trinomial.

4) A polynomial with only the constant term is referred to as a constant polynomial.

5) Polynomial - An expression of this type is a polynomial.

An expression with three variables is xy + yz + zx.

XY + YZ + ZX is a polynomial, hence.

Explanation: 

In the algebraic statement (i), the degree of a polynomial in one variable equals the largest power of the variable. 2x – 1 1 power of x

The expression's highest power for the variable x is 1 Consequently, the polynomial's degree 2x – 1 = 1 (ii) –10

The specified term contains no variables.

Suppose that x is a parameter in the provided statement.

– 10 = –10x0

Power of x equals zero

Therefore, grade of the polynomial - 10 = 0 (iii) Highest power of the variable x in the provided formula = 0. x3 – 9x + 3x5

Powers of x are 3, 1, and 5 in that order.

In the given opinion, the value of x's highest power is equal to 5. Consequently, the polynomial's degree is x3 - 9x + 3x5= 5 (i) y3 (1 - y4).

The formula may be expressed as y3 (1 - y4) = y3 - y7, where y is a factor of 3 while y is a value of 7.

Explanation:

The value of the polynomial is 15x3+2x5+1572x2x6 (x3+2x+⅕ 7/2 x²-x6).

(i) The polynomial has a value of 6, which is the highest power of the parameter.

(ii)In the polynomial in question, the x3 coefficient is 15.

In the above polynomial, the coefficient of x6 is -1.

(iv) The polynomial in question has a constant term of 15.

Explanation: 

Determining the coefficient: A polynomial's coefficient is a multiplicative factor.

The formula for 3x-5 is 0x²+3x5.

In the above polynomial, the slope of x2 is zero.

As a result, the factor of x² in the polynomials presented is 0.

Explanation: 

(i) Since x has a maximum value of 3, the polynomial 2-x²+x3 is a cubic polynomial.

(ii) Since 3 is x's largest exponent, the polynomial 3x3 is a cubic polynomial.

(iii) Since t has a maximum exponent of 1, polynomial 5t7 is a linear polynomial.

(iv) Due to the fact that y has a maximum exponent of 2, the polynomial 4-5y² is a quadratic polynomial.

Because the exponent of the variable is 0, the polynomial 3x0 is a constant polynomial (v).

(vi) Due to the fact that x has the largest exponent of 1, the polynomial 2+x is a linear polynomial.

(vii) Because y's largest exponent is 3, the polynomial y3y is a cubic polynomial.

(ix) The polynomial t² has the maximum exponent of 2 and is therefore a quadratic polynomial.

(x) The linear polynomial 2x1 is 2.

Explanation: 

(i) 5y or 10x are two examples of degree 1 monomials.

(ii) 6x20

+x11 or x20

+1 is an example of a binomial of degree 20.

(iii) x2

5x+4 or 2x2

x1 are examples of degree two trinomials.

Explanation: 

The polynomial is 3x - 4x2 + 7x - 5, given.

We must determine the polynomial's value at x = 3 and x = -3.

Let p(x) = x³ - x4 + x7 - x5 and p(3) = x3³ - 4(3)² + 7(3) - 5 = 3(27) - 4(9) + 21 - 5 = 81 + 21 - 36 - 5 = 102 - 41 = 61

The polynomial's value consequently equals 61 at x = 3.

p(-3) = 3(-3)³ - 4(-3)² + 7(-3) - 5 = 3(-27) - 4(9) - 21 - 5 = -81 - 36 - 26 = -81 - 62 = -143

As a result, the polynomial has a value of -143 at the value of x = 3.

Explanation: 

If p(x)=x²4x+3, then

presently, p(2)=(2)²−4×2+3=4−8+3=−1

 p(−1)=(−1)²−4(−1)+3=1+4+3=8

Furthermore p(1/2)=(1/2)²−4×1/2+3

 =14−2+3=1−8+12/4=5/4

∴ p(2)−p(−1)+p(12)=−1−8+5/4

 =−9+5/4=−36+5/4 =−31/4.

Explanation: 

The polynomial can be calculated as p(x) = 10x - 4x2 - 3.

Finding p(0), p(1), and p(-2) is necessary.

p(0) = 10(0) - 4(0)² - 3

= 0 - 3

p(0) = -3

As a result, the polynomial's value at p(0) is 3.

p(1) = 10(1) - 4(1)² - 3

= 10 - 4 - 3

= 10 - 7

p(1) = 3

Consequently, the polynomial's result at p(1) is 3.

p(-2) = 10(-2) - 4(-2)² - 3

= -20 - 4(4) - 3

= -20 - 16 - 3

= -36 - 3

p(-2) = -39

As a result, the polynomial's result at p(-2) is -39.

Explanation: 

(i) The zero of x -3 is -3.

False Zero of x-3 is produced by, x-3 = 0

=>x=3

(ii) - 1/3 is a zero of 3x+1. The formula for 3x + 1's true zero is 3x + 1 = 0 

=>3x = - 1

x = - 1/3

(iii). -4/5 represents a zero of 4 -5y.

 The formula 4 - 5y =0 - 

=>5y = - 4

=>y = 4/5 (iv) yields the false zero of 4 - 5y. The zeroes of t2 - 2t are 0 and 2. True Zeros of t2 - 2t are given by, for example, 2 (v) -3 is a zero of y2 + y - 6 or t2 - 2t = t(t - 2) = 0 t = 0. The formula y2 + y - 6 = 0

=>y2 + 3x - 2x - 6 = 0 

y (y + 3) - 2(x + 3) yields the true zero of the equation.

Explanation: 

The polynomial is expressed as g(x) = 3 - 6x.

The polynomial's zeros must be located.

The polynomial will come out to zero for some values of the variable. These numbers are referred to as polynomial zeros.

Let g(x) = 0 

3 - 6x = 0

3 = 6x

x = 3/6

x = 1/2 to determine the polynomial's zeros.

Explanation: 

P(x)=(x−2)²−(x−2)²

That is shown as a²-b².

P(x)=(x−2+x+2)(x−2−(x+2))

P(x)=2x(x−2−x−2)

P(x)=2x(−4)

P(x)=0

−8x=0

x=0

Consequently, x=0 is the polynomial P(x) zero.

Explanation: 

Because x4 + 1 is the primary polynomial

x - 1 makes up the remaining polynomial.

When the first polynomial is divided by the second, we must determine the ratio and the balance via true division.

The formula for the first polynomial is x4 + 0x³ + 0x² + x + 1.

In terms of division,

As a result, the amount left is 2, and the ratio is x3 + x2 + x + 1.

Explanation: 

According to the last theorem, if a polynomial p(x) is divided by a linear factor x-a, the result is p(a).

Consequently, p(x)=x³-2x²-4x-1 and g(x)=x+1

As a result, g(x)=0 1

The rest theorem application

The rest after dividing p(x) by g(x) equals P(-1) P(-1)=-13-2-12-4-1-1 = 0.

Consequently, the remaining.

Explanation: 

Assuming that p(x) = x³ - 5x² + 4x - 3

Moreover, g(x) = x - 2

It is necessary to determine that p(x) is a multiple of g(x).

We are aware that g(x) must be divisible by p(x) if p(x) is a multiple of g(x).

Let g(x) equal 0 x - 2

x = 2

Place x = 2 in p(x),

where p(2) = (2)³ - 5(2).² + 4(2) - 3

 = 8 - 5(4) + 8 - 3

 = 8 - 20 + 5

 = 13 - 20

 = -7 

p(x) ≠ 0

G(x) cannot be divided by p(x) since the remainder is not zero.

p(x) is not a multiple of g(x), hence this is true.

Explanation: 

Enter the polynomial p(x) = 69 + 11x - x2 + x3 as the input.

Let g(x) = x + 3 be the coefficient.

We must determine whether 69 + 11x - x2 + x3 is a factor of x + 3.

If g(x) is set to 0 x + 3 or 0 x = -3

Replace the value of x in p(x) with -3; 

p(-3) = 69 + 11(-3) - (-3)² + (-3)³

 = 69 - 33 - 9 - 27 

= 36 - 9 - 27 

= 9 - 9

 = 0

Since p(x) equals 0 when x = -3, x + 3 is p(x)'s component.

The factor of 69 + 11x - x2 + x3 is therefore x + 3.

Explanation: 

The polynomial can be calculated as p(x) = 3x² + 6x - 24.

The question is whether x - 2 is a factor of p(x).

g(x) = x - 2 is an example.

Letting g(x) equal 0 x - 2 x = 2

Replace x = 2 in p(x), p(2) = 3(2)², 6(2)-24 

= 3, 12-24= 12,

24-24 = 0, and p(2) = 0.

Since p(x) equals 0 at x = 2, x - 2 is p(x)'s factor.

As a result, the factor of 3x² + 6x - 24 is x - 2.

Explanation: 

If f(p)=p10-1 and g(p)=p11-1, then

The result of putting p = 1 in f(p) is f(1)=110=1=1=1=0.

Since (p - 1) is a factor of (p10-1) according to the factor theorem,

We now obtain g(1)=111=1=1=1=0 by setting p = 1 in g(p).

Theoretically, (p - 1) is a factor of (p11-1) because of this.

Explanation: 

The polynomial can be calculated as p(x) = x³ - 2mx² + 16.

If p(x) is divisible by x + 2, we need to determine the value of m.

Suppose g(x) = x + 2

Since g(x) divides p(x),

Consequently, g(x) = 0 x + 2 = 0 x = -2

Replace x = -2 with p(x),

Because p(x) and g(x) are split, p(-2) = 0 p(-2) = x3 - 2mx2 + 16 = 0 

(-2)³ - 2m(-2)² + 16 = 0 

-8 - 2m(4) + 16 = 0 

-8m + 8 = 0

 8m = 8 

m = 1

As a result, m = 1.

Explanation: 

The polynomial can be calculated as x⁵ - 4a²x3 + 2x + 2a + 3.

A factor of the polynomial is x + 2a.

We must ascertain what an is worth.

Allow p(x) to equal x⁵ - 4a²x³ + 2x + 2a + 3.

Suppose g(x) = x + 2a.

So, g(x) = 0

x + 2a = 0

x = -2a

Fill in p(x) with x = -2a;

Assumedly, g(x) is a component of p(x).

x5 - 4a2x

³ + 2x + 2a + 3 = p(-2a) = 0

(-2a)⁵ - 4a²(-2a)³ + 2(-2a) + 2a + 3 = 0

-32a⁵ - 4a²(-8a³) - 4a + 2a + 3 = 0

-32a⁵ + 32a⁵ - 2a + 3 = 0

-2a + 3 = 0

-2a = -3

a = 3/2

As a result, a has a value of 3/2.

Explanation: 

Take f(x)=8x4, 4x4, 16x², 10x, and m.

The fact that (2x-1) is a factor of f(x) is established.

We obtain f(1/2)=0 =>8x1/24+4x1/24-16x1/2²+10x12+m=0

=>12+12/4+5+m=0

=>2+m=0=m

=>2 applying the coefficient theorem.

Consequently, m has a value of 2.

Explanation: 

Let p(x) = ax³ + x²- 2 x + 4 a 9

Given that p(x) is a factor of x + 1, the rest of the theorem states that p(-1) = 0.

A(1)³+(-1)²-2(-1)+4a-9=0; 

=>A+1+2+4a9=0;

⇒ 3a–6=0⇒ 3a=6

=>a=6/3=2;

Consequently, a has an amount of 2.

Explanation: 

(i) 2 + 9 + 18 = 2 + 6 + 3 + 18 [By dividing the intermediate period]

=x(x+6)+3(x+6)=(x+3)(x+6)

(ii) 6x² + 7x 3 = 6x² + 9x ² + 3 [By dividing the intermediate period]

=3x(2x+3)–1(2x+3)=(3x–1)(2x+3)

(iii) 2x²-7x-15=2x²–10x+3x–15 [By dividing the intermediate period]

=2x(x–5)+3(x–5)=(2x+3)(x–5)

(iv) 84-2r-2r² = 2 (r² + r-42)

=−2(r²+7r–6r–42) [By dividing the intermediate period]

=−2[r(r+7)–6(r+7)]

=−2(r–6)(r+7)=2(6–r)(r+7).

Explanation: 

Let p(x) = 2x³-3x²-17x+30 (i).

P(x) constant term of 30

∴ The value of 30 is ±1,±2,±3,±5,±6,±10,±15,±30  30.

Trial and error revealed that p(2) = 0, indicating that (x - 2) is a factor of p(x).

[∵2(2)³−3(2)2−17(2)+30=16−12−34+30=0]

Currently, we can observe that 2x³-3x2-17x+30

=(x−2)(2x²+x−15)

2x²+x−15=2x(x+3)–5(x+3) [By dividing the intermediate period]

=(x+3)(2x–5)

x = (x+2)(x+3)(2x+5) when 2x³-3x²-17x+30

(ii) Suppose p(x)=x³-6x²+11x-6.

P(x) constant term: - 6

1, 2, 3, and 6 are components of -6.

Trial and error revealed that p(1) = 0. Therefore, p(x) is a factor of (x - 1).

[∵ (1)3−6(1)2+11(1)−6=1−6+11−6=0]

In the present, x³-6x²+11x-6 =x³-x²-5x²+5x+6x-6 =(x1)(x²-5x+6) [Using the standard factor (x - 1)]

Therefore, (x²-5x+6)=x²3x²x+6. [By dividing the intermediate period]

=x(x–3)–2(x–3) =(x–3)(x–2)

(x-1)(x-2)(x-3) = x³-6x²+11x-6

(iii) Suppose p(x)=x³+x²4x4.

P(x) fixed term: - 4

1, 2, and 4 are the -4 factors.

Trial and error revealed that p(-1) = 0. Therefore, p(x) is a factor of (x + 1).

Now, x³+x²−4x−4 =x²(x+1)–4(x+1) =(x+1)(x²–4) [Using (x).

Explanation: 

(i) 103³=(100+3)³

 =(100)³+(3)³+3×100×3(100+3) Inferring from the belonging, (a+b)³=a³+b³+3ab(a+b)

=1000000+27+900(103)

 =1000027+92700=1092727

(ii) 101×102=(100+1)(100+2)

 =(100)²+100(1+2)+1×2 

Inferring using the belonging, (x+a)(x+b)=x²+x(a+b)+ab

=10000+300+2=10302

(iii) (999)²=(1000−1)² =(1000)²+(1)²–2×1000×1 (a-b)²=a²+b²-2ab [using the similarity]

=1000000+1–2000=998001.

Explanation: 

(i): 4x²+20x+25=(2x)²+2×5×2x+(5)² =(2x+5)A²+2ab+b2=(a+b)² The use of the same]

(ii) 9y² 66yz plus 121z² equals(3y)²–2×3y×11z+(11z)² =(3y–11z)² A2-2ab+b2=(a-b)² [which utilised the identity]

(iii) (2x+1/3)²(x−1/2)²=[(2x+1/3)−(x−1/2)][(2x+1/3)+(x−1/2)]

A²-b²=(a-b)(a+b) [which utilised the identity]

=(2x−x+1/3+1/2)(2x+x+1/3−1/2)

 =(x+2+3/6)(3x+2−3/6) 

=(x+5/6)(3x−1/6).

Explanation: 

(i) 9x²–12x+3=3

(3x²–4x+1) [using 3 as a baseline]

=3(3x²–3x–x+1) [By blowing out the interjection]

=3[3x(x−1)–1(x–1)]=3[(3x−1)(x–1)]

(ii) 9x²-12x+4=(3x)²–2×3x×2+(2)²

 =(3x−2)² Inferring from the sense of self, (ab)²=a²-2ab+b²

=(3x-2)(3x–2).

Explanation: 

(i) (4a−b+2c)²=(4a²)+(−b)²+(2c)²+2(4a)(−b)+2(−b)(2c)+2(2c)(4a)

Utilising the fact that (a+b+c)² = a² + b + c + 2 a + b + 2 c + 2 a]

=16a²+b²+4c²–8ab–4bc+16ac

(ii) (3a–5b–c)²=(3a)²+(−5b)²+(−c)²+2(3a)(−5b)+2(5b)(−c)+2(−c)(3a)

Utilising the formula: (a+b+c)2 = a² + b + c + 2 a + b + 2 c + 2 a]

=9a²+25b²+c²–30ab−10bc–6ac

(iii) (−x+2y−3z)²=(−x)²+(2y)²+(−3z)²+2(−x)(2y)+2(2y)(−3z)+2(−3z)(−x)

Utilising the formula: (a+b+c)² = a² + b + c + 2 a + b + 2 c + 2 a]

=x²+4y²+9z²–4xy–12yz+6xz.

Explanation: 

The formula is as follows: 9x² + 4y² + 16z² + 12xy - 16yz - 24xz - First, we have to consider the phrase.

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca is the mathematical equality. ---------- (2) Comparing (1) and (2), we see that a2 = 9x2

a = 3x,

b2 = 4y²

b = 2y,

and c2 = 16z2 

c = 4z.

2ab = 12xy,

ab = 6xy,

bc = -8yz,

2y(c) = -8yz,

c = -8yz, and 2y c = -4z.

Ca = -12xz and 2ca = -24xz.y/3

Therefore, a = 3x, b = 2y, and c = -4z.

At this point, (a + b + c)² = (3x + 2y - 4z).²

Hence, (3x + 2y - 4z) and (3x + 2y - 4z) are the variables that makeup of the provided formula.

Explanation:

If a + b + c = 9, then

Ab = bc + ca = 26 as well.

We must determine a² + b² + c².

(A + B + C) is a logical formula.2 = 2ab + 2bc + 2ca = a² + b² + c² ---------- (1)

Finally, (a + b + c)² = (9)² = 81

The formula for the problem (1) is (a + b + c)²

 = (ab + bc + ca)² + ² + b² + c²

So, 81 = a² + b² + c² + 2(26)

51 = a² + b² + c² + 52 51 - 52

a² + b² - c² = 29

As a result, a² + b² + c² = 29.

Explanation: 

(i) (3a–2b)³=(3a)³+(−2b)³+3(3a)(−2b)(3a–2b)

The use of (a,b) as an individual].3=a3–b3+3a(−b)(a–b)]

=27a³–8b³–18ab(3a–2b)

 =27a³–8b³–54a2b+36ab2

 =27a³–54a2b+36ab2−8b³

(ii) (1/x+y³)³=(1/x)³+(y³)³+3(1/x)(y³)(1/x+y³)

Inferring from the belonging, (a+b)³=a³+b³+3ab(a+b)

=1/x³+y³/27+y/x(1/x+y³)=1/x³+y³/27+y/x²+y/2³x

(iii) (4−1/3x)³=(4)³+(−1/3x)3+3(4)(−1/3x)(4−1/3x)

[By through the concept of identity, (ab)³=a³-b³+3a(ab)(a-b)]

=64−1/27x³−4/x(4−1/3x) 

=64−1/27x³−16/x+4/3x².

Explanation: 

The calculation is 1 - 64a³ - 12a + 48a², which is presented. First, we need to consider the phrase.

Making use of the algebraic belonging,

The equation (a - b)³ is (a - b)³ = a³ - b³ - 3a²b + 3ab². ------------- (2)

When (1) and (2) are compared, a3 = 1

 a = 1 

b³ = 64a³.

(1) b = 4a,

 -3a2b = 12, 

and a²b = -4a²

b = -4a

Thus, b = -4a,

 3ab² = 48a²,

 ab² = 16a²,

 and b² = 4a.

Here, a = 1 and b = -4a.

So, (a - b)³ = (1 - 4a)³ 

Thus, (1 - 4a), (1 - 4a), and (1 - 4a) are the variables of the provided equation.

Explanation: 

The formula for this is (x/2 + 2y)(x2/4 - xy + 4y²), which is given.

We must identify the expression's output.

(x/2 + 2y)(x2/4 - xy + 4y²) equals (x/2 + 2y)(x2/4 - xy + 4y²).

By virtue of the distributive and exponential properties, x/2(x2/4 - xy + 4y²) = x/2(x2/4) x2y/2 + 4xy2/2 = x³/8 x2y/2 + 2xy² 

= x³/8 x2y/2 + 4xy2/2

By virtue of the multiplying and dividing properties, 2y(x2/4 - xy + 4y2) = 2y(x2/4) - xy + 4y² 

= 2x2y/4 - xy² + 8y³ 

= x2y/2 - xy² + 8y³.

In other words, (x/2 + 2y)(x2/4 - xy + 4y²) = x³/8 - x2y/2 + 2xy² + x2y/2 - 2xy² + 8y³.

By arranging, the formula is x3/8 + x2y/2 - x2y/2 - 2xy² + 2xy² + 8y3 = x³/8 + 8y³.

Consequently, the expression's result is x3/8 + 8y³.

Explanation: 

(i) 1+64x3=(1)3+(4x)3 =(1+4x)[(1)2–(1)(4x)+(4x)2]

Utilizing the fact that a²ab+b²=a³-b³]

=(1+4x)(1–4x+16x²)

(ii) a³−2√2b³=(a)³–(√2b)³ =(a−√2b)[a²+a(√2b)+(√2b)²]

A³-b³=(a-b)(a²+ab+b²) utilising the identity

(a−√2b)(a²+√2ab+2b²).

Explanation: 

(2x−y+3z)(4x²+y²+9z²+2xy+3yz−6xz) =(2x−y+3z)(2x)²+(−y)²+(3z)²−(2x)(−y)−(−y)(3z)−(2x)(3z) 

=(2x)³+(−y)³+(3z)³−3(2x)(−y)(3z)

Utilising the the same, a³+b³+c³3abc] (a+b+c)(a²+b²+c²abbcca)

=8x³−y³+27z³+18xyz.

Explanation: 

(i) a³-8-b-64-c-24-abc(a)³+(−2b)³+(−4c)³−3×(a)×(−2b)×(−4c) =(a−2b−4c)[(a)²+(−2b)²+(−4c)²−a(−2b)−(−2b)(−4c)−(−4c)(a)]

Applying the name, a²+b²+c² abbcca=(a+b+c)(a ³+b ³+c ³3abc=(a+b+c)(a+b+c)]

=(a−2b−4c)(a²+4b²+16c²+2ab−8bc+4ac)

(ii) 2 2a³+8 b³+27 c³+18 2abc = 2 a ³ + 2 b ³ + 3 c³−3(√2a)(2b)(−3c) =(√2a+2b−3c)[(√2a)²+(2b)²+(−3c)²−(√2a)(2b)−(2b)(−3c)−(−3c)(√2a)]

Applying the name, a²+b²+c²abbcca=(a+b+c)(a³+b³+c³3abc=(a+b+c)(a+b+c)]

=(√2a+2b−3c)[2a²+4b²+9c²−2√2ab+6bc+3√2ac].

Explanation: 

(0.2)³−(0.3)³+(0.1)³

 Since x³+y³+z³3xyz=(x+y+z)³ is known,

When x+y+z=0, x3+y3+z33xyz=0 follows.

⇒x³+y³+z³=3xyz

We can calculate x=0.2, y=0.3, and z=0.1 using the given data.

After that, x+y+z=0.20.3+0.1=0.

Consequently, (0.2)³+(−03)³+(0.1)³=3(0.2)(−0.3)(0.1)=−0.018

Explanation: 

Having said it,

a³+b³+c³−3abc=(a+b+c)(a²+b²+c²−ab−bc−ca)

Additionally, if a + b + c = 0, then a 3 + b 3 + c 3 = 3 ABC (condition 1)

We can see that a+b+c=(x2y)+(2y3z)+(3zx)=0 here.

The result of condition (1) is (x2y)³+(2y³z)³+(3zx)³ =3(x−2y)(2y−3z)(3z−x).

Explanation: 

(i) Assuming that x + y + 4 = 0, x 3 + y 3 + 4 = xy (4).(1) The use of the fact that a³+b³+c³=3abc if a+b+c=0]

=12xy

Now, x³+y³−12xy+64=x³+y³+64−12xy =12xy−12xy=0 (i) of Equation

Since x2y6=0,

∴ x³+(−2y)³+(−6)³=3x(−2y)(−6)

Utilizing the fact that a³+b³+c³=3abc if a+b+c=0]

x³−8y³−216=36xy …..(i)

Currently, x³8y³216³6xy =x³8y³216³6xy =36xy³6xy=0. (i) Equation

Explanation: 

Provided, 4a² + 4a - 3 gives the rectangle's area.

We need to identify all feasible formulas for the rectangle's width and height.

By dividing the centre term, 4a² + 4a -3 can be factored as 4a² + 6a - 2a - 3 = 2a(2a + 3) - 1(2a + 3) = (2a - 1)(2a + 3)

Consequently, 2a - 1(2a + 3) = 4a² + 4a -3.

Explanation: 

The polynomials are as follows: z³ - 4z + a and az³ + 4z2 + 3z - 4.

When divided by z – 3, the polynomials provide an identical amount of remainder.

So must ascertain what an is worth.

Let g(x) = z - 3 and p(x) = az³ + 4z2 + 3z - 4

Currently, g(x) = 0 z - 3 = 0 z = 3.

Insert z = 3 into p(x), where p(3) = a(3)3 + 4(3)² + 3(3) - 4 

= 27a + 4(9) + 9 - 4

 = 27a + 36 + 5 

= 27a + 41

Allow q(x)=z³ - 4z + a.

Using the formula z = 3 in q(x), q(3) = (3)³ - 4(3) + a 

= 27 - 12 + a

 = 15 + a

Considering that p(x)/g(x) = q(x)/g(x) has a residual,

So,

 27a + 41 = 15 + a

27a - a = 15 - 41

26a = -26 a = -1

As a result, a has a score of 1.

Explanation: 

Considering that the polynomial is p(x) = x⁴ - 2x³ + 3x² - axe + 3a - 7, p(x) divides by x + 1 to remove the remainder.

If p(x) has been divided by x + 2,

 we must determine the value of a and the remaining fraction.

Suppose g(x) = x + 1 g(x) = 0 x + 1 x = -1

Assuming x = -1, make p(x) = p(-1)⁴ - 2(-1)³ + 3(-1)² - a(-1) + 3a - 7

 = 1 -2(-1) + 3 + a + 3a - 7 

= 1 + 2 + 3 + 4a - 7

 = 6 - 7 + 4a 

= 4a - 1

p(-1)/g(-1) = 19 provided

So, 4a - 1 = 19

4a = 19 + 1

 4a = 20

 a = 20/4

 a = 5

Consequently, a has a score of 5.

P(x) equals x⁴ - 2x³ + 3x² - 5x + 3(5) - 7

 and P(x) equals x⁴ - 2x³ + 3x² - 5x + 8

Suppose q(x) = x + 2

At this point, q(x) = 0 x + 2 

= 0 x = -2

Include x = -2 in p(x),

making p(-2) = (-2)⁴ - 2(-2)³ + 3(-2)² - 5(-2) + 8

 = 16 -2(-8) + 3(4) + 10 + 8

 = 16 + 16 + 12 + 18

 = 32 + 30 

= 62

Thus, when p(x) is divided by x + 2, 62 is the remaining.

Explanation: 

Given that the polynomial has the elements px² + 5x + r x - 2 and x - 1/2, it can be written as follows.

We must demonstrate that p equals r.

Let f(x) = px², 5x, and r.

g(x) = x - 2 g(x) = 0 x - 2 = 0 x = 2 at this point.

Using the formula x = 2 in p(x), p(2) = p(2)² + 5(2) + r = 4p + 10 + r

At x = 2, p(x) = 0

So, 4p + 10 + r = 0 —-------------- (1)

So, h(x) = x - 1/2 now

Put x = 1/2 in p(x), and the equation becomes p(1/2) = p(1/2)² + 5(1/2) + r

= p/4 + 5/2 + r

= (p + 2(5) + 4r)/4

= (p + 10 + 4r)/4.

When x equals to half, p(x) = 0 (p + 10 + 4r)/4 = 0 (p + 10 + 4r = 0). —------------ (2)

When (1) and (2) are compared, 4p + 10 + r equals p + 10 + 4r.

Grouping results in 4p - p + 10 - 10 = 4r - r.

3p = 3r p = r

As a result, it is established that p = r.

Explanation: 

As a result, it is established that p = r.

The polynomial is 2x⁴ - 5x³ + 2x² - x + 2 provided.

While actually dividing the polynomial, we must demonstrate that it is divisible by x² - 3x + 2.

The factor is as follows: x2 - 3x + 2.

After splitting the middle term to calculate x² - 3x + 2, the result is x(x - 2) - 1(x - 2) = (x - 1)(x - 2)

The variables are therefore (x - 1) and (x - 2).

Make p(x) equal to 2x⁴ - 5x³ + 2x² - x + 2

Let q(x) equal x- 1

Knowing that q(x)=0x-1=0x=1

Assuming that x = 1 

in p(x), p(1) = 2(1)⁴ - 5(1)³ + 2(1)² - (1) + 2 

= 2 - 5 + 2 - 1 + 2 

= 2 + 2 + 2 - 5 - 1

 = 6 - 6 

p(1) = 0

Let r(x) equal x- 2

Knowing that r(x)=0

Let x = 2 in p(x), and p(2) = 2(2) as x = 2⁴ - 5(2)³ + 2(2)² - (2) + 2 

= 2(16) - 5(8) + 2(4) - 2 + 2 

= 32 - 40 + 8 

= 32 + 8 - 40 

= 40 - 40 

p(2) = 0

Consequently, p(1) = p(2) = 0.

Explanation: 

Assuming, (2x - 5y)³ - (2x + 5y)³

The supplied expression needs to be made simpler.

The algebraic identity (a + b) is used.3 equals (a + b)3ab(a + b).

Here, b = 5y (2x + 5y) and a = 2x.³ 

= (2x)³ + (5y)8x³ + 125y³ + 30xy(2x + 5y) (2x + 5y)

 = 3 + 3(2x)(5y)(2x + 5y)³

 = 8x3, 125y³, 60x2y, and 150xy².

(A - B) is the algebraic identity.3 = a(-b)(a - b)(a - b)(a - b)

In this case, a = 2x and b = (5y - 2x)³ = (2x)³ - (5y)8x³ - 125y³ - 30xy(2x - 5y)(2x + 5y) = 3 + 3(2x)(-5y)(2x - 5y)³

= 8x3, 125y3, 60x²y, and 150xy².

Now, (2x - 5y)³ - (2x + 5y)3 is equal to 8x3 - 125y³ - 60x²y + 150xy² - (8x³ + 125y³ + 60x²y + 150xy2) 

= 8x³ - 125y³ - 60x²y - 150xy².

= - 250y³ - 120x2y 

= - 125y³ - 125y³ - 60xy - 60x²y

As a result, 250y³ - 120x2y = (2x - 5y)³ - (2x + 5y)³.

Explanation: 

The formula is as follows: x ²+ 4y² + z² + 2xy + xz - 2yz

The expression must be multiplied by (-z + x - 2y).

(-z + x - 2y) = (-z(x² + 4y² + z² + 2xy + xz - 2yz) + x(x² + 4y² + ²2 + 2xy + xz - 2yz) (x2 + 4y²+ z² + 2xy + xz - 2yz) - 2y

In other words, -z(x²+ 4y² + z² + 2xy + xz - 2yz) 

= x² (-z) + 4y² (-z) + z² (-z) + 2xy (-z) + xz (-z) - 2yz (-z) = -zx² - 4y²z - z³ - 2xyz. - xz² + 2yz²

In other words, x(x² + 4y² + z² + 2xy + xz - 2yz) 

= x² + 4y² + z² + 2xy + xz x3 + 4xy² + xz² + 2x²y + x²z - 2xyz = 2yz(x)

2x²y - 8y³ - 2yz² - 4xy² - 2xyz + 4y²z

= -2y(x² + 4y² + z² + 2xy + xz - 2yz) 

= x²(-2y) + 4y²(-2y) + z²(-2y) + 2xy(-2y) + xz(-2y) - 2yz(-2y).

As a result, (-z + x - 2y) (x² + 4y² + z² + 2xy + xz - 2yz) 

= -zx²- 4y²z - z3 - 2xyz - xz2²+ 2yz² + x3 + 4xy² + xz² + 2x²y + x2z - 2xyz - 2x²y

8y³ + 2yz² + 4xy² + 2xyz + 4y²z + 2xyz + 2xyz

By combining,

= - z³ + x³ - 8y³ + 2x²y - 2x²y - 4xy², 4xy², 4y²z, 2yz², 2yz², 4y²z, x²z, xz², 2xyz, 2xyz, and 2xyz.

= - Z³ + X³ - 8Y³ - 2XYZ - 2XYZ

= - z³+x³-8y³- 6xyz

In light of this, (-z + x - 2y) (x² + 4y² + z² + 2xy + xz - 2yz) = x³ - 8y³ - z³ - 6xyz.

Explanation: 

A, B, and C are all non-zero given that

A + B + C = 0 as well.

It must be demonstrated that a²/bc + b²/ca + c²/ab equals 3.

x³ + y³ + z³ - 3xyz using the algebraic identity equals (x + y + z) (x² + y² + z² - xy - yz - zx)

x³ + y³ + z³ - 3xyz equals 0 if x + y³ + z = 0.

Thus, 3xyz = x³ + y³ + z³.

If a + b + c = 0, then

A³ + B³ + C³ equals 3ABC.

A³/3abc + B³/3abc + C³/3abc = 3abc/3abc when all sides are divided by 3abc.

A²/3bc = 1 and A²/3bc + B²/3ca + C2/3ab = 3

So it was proved.

Explanation: 

Since a+b+c=5, this proves that a+b+c=3abc=25.

(5)2=a²+b²+c²+2(ab+bc+ca) where ab+bc+ca=10 and (a+b+c)2

=a²+b²+c²+2(ab+bc+ca)(10), 

a²+b²+c²=25-20, 

25=a²+b²+c²+20

⇒ a²+b²+c²=5

This is demonstrated by the equation LHS=a³+b³+c³3abc =(a+b+c)(a²+b²+c²ab+bc+ca)

=(5)[5(ab+bc+ca)] =5(5-10)=5(5)=25=RHS.

Explanation: 

Proof: (a+b+c)³-a³-b³-c³=3(a+b)(b+c)(c+a).

LHS=[(a+b+c)³–a³]–(b³+c³) =(a+b+c−a)[(a+b+c)²+a²+a(a+b+c)]−[(b+c)(b²+c²−bc)]

A³+b³=(a+b)(a²+b²ab) and a³-b³=(ab)(a²+b²+ab) are identities.

=(b+c)[a²+b²+c²+2ab+2bc+2ca+a²+a²+ab+ac]−(b+c)(b²+c²−bc) =(b+c)[b²+c²+3a²+3ab+3ac−b²−c²+3bc]

=(b+c)[3(a²+ab+ac+bc)]

=3(b+c)[a(a+b)+c(a+b)]

=3(b+c)[(a+c)(a+b)]

=3(a+b)(b+c)(c+a)=RHS Thus, the proof.