Chapter-2, Polynomials

Explanation:

A graphic way to detect zeroes is to compare all times the curve crosses the x-axis in any polynomial equation with all of its zeros.

(i) The graph is parallel to the x-axis and does not cross it at any point, the number of zeros in p(x) in the graph equals zero.

(ii) Because the graph only has one place where it crosses the x-axis, the number of zeros in p(x) is 1 in the graph.

(iii) P(x) in this case contains three zeros because the graph contacts the x-axis three times.

(iv) The graph's p(x) has two zeros since it touches the x-axis twice.

(v) In the supplied graph, p(x) contains four zeros because the graph touches the x-axis four times.

(vi) In the example, the graph touches the x-axis three times, and p(x) contains three zeros.

Explanation:

(i) x2–2x –8

x2- 4x+2x-8 equals x(x-4)+2(x-4)=(x-4)(x+2).

Hence, the x2-2x-8 polynomial equation’s zeros are (4,-2)

The sum of zeroes is equal to = 4–2 = 2 = -(-2)/1 = - (Coefficient of x)/. (Coefficient of x2)

Moreover, the product of zeroes is 4(-2) = -8=(-8)/1 (constant term)/. (Coefficient of x2)

(ii) 4s2–4s+1

⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Hence, 4s2–4s+1 polynomial equation’s zeros are (1/2, 1/2)

The Sum of zeroes is equal to  (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s2)

And Product of zeros is equal to  (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 )

(iii) 6x2–3–7x

⇒6x2–7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)

Hence , 6x2–3–7x  polynomial equation’s zeros are (-1/3, 3/2)

The Sum of zeroes is equal to  -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2)

 The Product of zeroes is equal to  -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x2)

(iv) 4u2+8u

⇒ 4u(u+2)

Hence, the 4u2 + 8u polynomial equation’s zeros are (0, -2).

The  Sum of zeroes is equal to  0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2)

The Product of zeroes is equal to  0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 )

(v) t2–15

⇒ t2 = 15 or t = ±√15

Hence  t2 –15  polynomial equation’s zeros are (√15, -√15)

The Sum of zeroes is equal to √15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2)

The  Product of zeroes  is equal to  √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 )

(vi) 3x2–x–4

⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Hence, 3x2 – x – 4 polynomial equation’s zeros are (4/3, -1)

The Sum of zeroes is equal to (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2)

The Product of zeroes  is equal to (4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x2 ).

(i) 1/4, -1

Explanation:

The sum and product of zeroes formulas

Total of zeros =  α+β

The product of zeros is α β

Then zeroes sum = α + β = 1/4

Here, the product of zeroes is -1. In this case, the quadratic polynomial equation can be written as follows if and are zeros in any quadratic polynomial:

x2–(α+β)x +αβ = 0

x2–(1/4)x +(-1) = 0

4x2–x-4 = 0

Therefore  4x2–x–4 is the quadratic polynomial.

(ii)√2, 1/3

Explanation:

Total of zeroes = α + β =√2

And Product of zeroes( α β) is ⅓

 In this case, the quadratic polynomial equation can be written as follows if and are zeros in any quadratic polynomial:

x2–(α+β)x +αβ = 0

x2 –(√2)x + (1/3) = 0

3x2-3√2x+1 = 0

Therefore, 3x2-3√2x+1 is the quadratic polynomial.

(iii) 0, √5

Explanation:

In the Given,

Total of zeroes = α+β = 0

The product of zeroes (α β) is  √5

 In this case, the quadratic polynomial equation can be written as follows if and are zeros in any quadratic polynomial:

x2–(α+β)x +αβ = 0

x2–(0)x +√5= 0

Therefore  x2+√5 is the quadratic polynomial

(iv) 1, 1

Explanation:

Total of zeroes = α+β = 1

The product of zeroes (α β) is  1

In this case, the quadratic polynomial equation can be written as follows if and are zeros in any quadratic polynomial:

x2–(α+β)x +αβ = 0

x2–x+1 = 0

Therefore, x2–x+1 is the quadratic polynomial.

(v) -1/4, 1/4

Explanation:

Total of zeroes = α+β = -1/4

The product of zeroes (α β) is  1/4

In this case, the quadratic polynomial equation can be written as follows if and are zeros in any quadratic polynomial:

x2–(α+β)x +αβ = 0

x2–(-1/4)x +(1/4) = 0

4x2+x+1 = 0

∴ Therefore, 4x2+x+1 is the quadratic polynomial

(vi) 4, 1

Explanation:

The Total of zeroes = α+β =4

The Product of zeroes (αβ)  is  1

In this case, the quadratic polynomial equation can be written as follows if and are zeros in any quadratic polynomial:

x2–(α+β)x+αβ = 0

x2–4x+1 = 0

∴ Therefore,x2–4x+1 is the quadratic polynomial.

(i) p(x) = x3-3x2+5x–3 , g(x) = x2–2

Explanation:

From the given

Dividend = p(x) = x3-3x2+5x–3

Divisor = g(x) = x2– 2

Here, after division, we obtain,

Quotient = x–3

Remainder = 7x–9

(ii) p(x) = x4-3x2+4x+5 , g(x) = x2+1-x

Explanation:

From the Given

Dividend = p(x) = x4 – 3x2 + 4x +5

Divisor = g(x) = x2 +1-x

Here, after division, we obtain,

Quotient = x2 + x–3

Remainder = 8

(iii) p(x) =x4–5x+6, g(x) = 2–x2

Explanation:

From the given

Dividend = p(x) =x4 – 5x + 6 = x4 +0x2–5x+6

Divisor = g(x) = 2–x2 = –x2+2

Here, after division, we obtain,

Quotient = -x2-2

Remainder = -5x + 10.

(i) t2-3, 2t4 +3t3-2t2-9t-12

Explanation:

As Given,

The First polynomial is equal to  t2-3

And the Second polynomial is equal to  2t4 +3t3-2t2 -9t-12

Here, the remaining value is set to 0. Hence, we declare that t2-3 is a factor of 2t4 + 3t3-2t2 -9t-12.

(ii)x2+3x+1 , 3x4+5x3-7x2+2x+2

Explanation:

As Given,

The First polynomial is equal to  x2+3x+1

The Second polynomial is equal to  3x4+5x3-7x2+2x+2

 0 is left as the final result. Therefore we argue that x2 + 3x + 1 is a factor of 3x4 + 5x3 - 7x2 + 2x+ 2.

(iii) x3-3x+1, x5-4x3+x2+3x+1

Explanation:

As Given,

The First polynomial is equal to  x3-3x+1

The Second polynomial is equal to  x5-4x3+x2+3x+1

The remainder does not add up to zero. As x3-3x+1 is not a factor of x5-4x3+x2+3x+1, we state that.

Explanation:

There will be a total of 4 roots in a polynomial equation of degree 4.

The polynomials f's zeros are (5/3) and - (5/3) (x).

A factor of the given polynomial f is (x -(5/3)) (x+(5/3) = x2-(5/3) = 0 (3x2-5)=0 (x).

The result of this division, f(x), by (3x25) is a factor of f(x), and the remainder is 0.

Explanation:

Given,

Dividend, p(x) = x3-3x2+x+2

Quotient = x-2

Remainder = –2x+4

We must now determine the value of the divisor, g(x) =?

Dividend = Divisor × Quotient + Remainder

∴ x3-3x2+x+2 = g(x)×(x-2) + (-2x+4)

x3-3x2+x+2-(-2x+4) = g(x)×(x-2)

Hence, g(x) × (x-2) = x3-3x2+3x-2

We will divide x3-3x2+3x-2 with the result of obtaining g(x) and (x-2)

Hence, g(x) = (x2–x+1)

Thus, 3x4 +6x3 −2x2 −10x–5 = (3x2 –5)(x2+2x+1)

(x2+2x+1) on additional factoring, we have,

x2+2x+1 = x2+x+x+1 = 0

x(x+1)+1(x+1) = 0

(x+1)(x+1) = 0

Both x=1 and x=1 are used to give zeroes.

The polynomial equation's four zeros are as follows:

√(5/3),- √(5/3) , −1 and −1.This is a solution.

Explanation:

According to the division method, the dividend p(x) and the divisor g(x) are two polynomials, with g(x)0. The values of the quotient q(x) and remainder r(x) can be obtained using the following formula: Dividend = Divisor × Quotient + Remainder p(x) = g(x)q(x)+r (x)

Where r(x) = 0 or r(x) degree > g degree (x).

We will now use examples to demonstrate how the three scenarios described above can be divided.

(i) deg p(x) = deg q(x)

The degree of the dividend and the degree of the quotient are only equal when the divisor is a constant term.

Now let's look at an example. The polynomial p(x) = 3x2+3x+3 can be divided by the integer g(x) = 3.

Hence, (3x2+3x+3)/3 = x2+x+1 = q (x)

Thus, we can see that the degree of the quotient q(x) = 2 and the degree of the dividend p are both equal (x).

The division algorithm is met in this situation.

(ii) deg q(x) = deg r(x)

We'll use p(x) = x2 + 3 as an example, a polynomial that can be divided by g(x) = x - 1.

So, x2 + 3 = (x – 1) (x – 1)

×(x) + (x + 3)

Q(x) = quotient, R(x) = remainder, X + 3

The degree of the quotient q(x) = 1 can be seen here, and it is also equal to the degree of the remainder r. (x).

The algorithm for division is satisfied in this case.

(iii) deg r(x) = 0

The degree of remainder equals 0 only when the remainder left behind after division is constant.

Let's use p(x) = x2 + 1 as an example of a polynomial that can be divided by g(x) = x.

Therefore, x2 + 1 = (x)×(x) + 1

The quantity q(x) equals x

Restatement r(x) = 1

The degree of residual, in this case, is 0.

Here, the division algorithm is successful.

(i) 2x3+x2-5x+2; -1/2, 1, -2

Explanation:

p(x) = 2x3+x2-5x+2 given

Therefore p(x) zeroes are = 1/2, 1, and -2. 

∴ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1)3+(1)2-5(1)+2 = 0

p(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0

The zeros of 2x3+x2-5x+2 are therefore demonstrated to be 1/2, 1, and -2.

The result of comparing the provided polynomial with the general equation is;

∴ ax3+bx2+cx+d = 2x3+x2-5x+2

a=2, b=1, c= -5 and d = 2

As we know, if the zeroes of the cubic polynomial ax3+bx2+cx+d are α, β,  and γ  then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Thus, by doing the polynomial's zero values,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

As a result, both the zeroes and coefficient values were satisfied.

(ii) x3-4x2+5x-2 ;2, 1, 1

Solution:

If p(x) = x3-4x2+5x-2, 

Then the p(x), the zeros are 2, 1, 2.

∴ p(2)= 23-4(2)2+5(2)-2 = 0

p(1) = 13-(4×12 )+(5×1)-2 = 0

So it is shown that the zeros of x3-4x2+5x-2 are 2, 1, and 1.

The result of comparing the provided polynomial with the general equation is;

∴ ax3+bx2+cx+d = x3-4x2+5x-2

a = 1, b = -4, c = 5 and d = -2

As we are aware, if the zeroes of the cubic polynomial ax3+bx2+cx+d are α, β,  and γ  then;

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

Thus, by setting the polynomial's zero values,

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

As a result, the zeroes and coefficients relationship is satisfied.

Explanation:

Let's assume that the cubic polynomial has the following values for its zeroes: α, β, γ.i.e ax3+bx2+cx+d.

In response to the query,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Hence, using the three formulas above, we can determine the polynomial coefficient values.

a = 1, b = -2, c = -7, d = 14

The cubic polynomial is therefore x3-2x2-7x+14.

Explanation:

The polynomial is provided to us here,

p(x) = x3-3x2+x+1

Zeroes are also given as a - b, a, and a + b.

The result of comparing the provided polynomial with the general equation is;

x3-3x2+x+1 = px3+qx2+rx+s

P, R, and S are all equal to one,q=-3

Zero-sum = a - b + a + a + b

-q/p = 3a

placing q and p as values.

-(-3)/1 = 3a

a=1

The zeroes are therefore 1+b, 1, and 1-b.

The zero product is now 1(1-b)(1+b).

-s/p = 1-b2

-1/1 = 1-b2

b2 = 1+1 = 2

b = ±√2

The zeros of x3-3x2+x+1 are therefore 1-√2, 1,1+√2.

Explanation:

As a result of this polynomial equation having a degree of 4, there will be a total of 4 roots.

Let f(x) equal x4-6x3-26x2+138x-35.

Given polynomial f (x) has zeros since 2 +√3 and 2-√3.

∴ [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

The result of multiplying the above equation is,

A factor of the given polynomial f is x2-4x+1 (x).

Now, if we divide f(x) by g(x), the residual will be zero and the quotient will also be a factor of f(x).

In other words, x4-6x3-26x2+138x-35 = (x2-4x+1)(x2 -2x-35).

Hence, when we factor (x2-2x-35) further, we obtain,

x2-(7−5)x −35 = x2- 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

Thus, its zeros come from:

x= −5 and x = 7.

As a result, the given polynomial equation's four zeros are: 2+√3, 2-√3, -5 and 7, respectively.

Explanation:

Take x4 - 6x3 + 16x2 - 25x + 10 and divide it by x2 - 2x + k.

Assuming that the polynomial division's final product is x + a.

(4k – 25 + 16 – 2k)

x + [10 – k(8 – k)] = x + a

x + a = (2k - 9)x + (10 - 8k + k2)

Comparing the aforementioned equation's coefficients, we discover;

2k – 9 = 1

2k = 9 + 1 = 10

k = 10/2 = 5

And,

10 – 8k + k2 = a

Given that k = 5, 10 - 8(5) + (5)2 Equals a.

10 – 40 + 25 = a

a = -5

As a result, a is -5 and k is 5.