Chapter-2, POLYNOMIALS NCERT Extra Questions

Explanation:

It is given in the above that, quadratic polynomial is (k-1) x² + k x + 1.

One zero of the polynomial is -3.

Now here we have to find the value of k.

Let f(x) = (k-1) x² + k x + 1

f(-3) = 0

Put x = -3 in the given polynomial

(k-1) (-3)² + k (-3) + 1 = 0

(k-1)(9) - 3k + 1 = 0

9k - 9 - 3k + 1 = 0

By grouping,

9k - 3k - 9 + 1 = 0

6k - 8 = 0

6k = 8

k = 8/6

k = 4/3

Therefore, the value of k is 4/3.

Explanation:

substitute x=-3 or 4 in the given options

a)x square - x + 12=0

9+3+12

not equals to 0

a is wrong

b)x square + x + 12

9-3+12

not equals zero

d)2x square+2x+24=0

x square + x+ 12=0

from b)

d is wrong

Therefore c is correct.

Explanation:

a=0,b=-6

Step-by-step explanation:

x=2

Hence, x²+(a+1)x+b=0

=>(2)²+(a+1)2+b=0

=> 4+2a+2+b=0

=> 6+2a+b =0

=> 2a+b =-6 ------------equation (1)

x=-3

Hence,x²+(a+1)x+b=0

=>(-3)²+[-3(a+1)]+b=0

=> 9+[-3a+(-3)]+b=0

=> 9-3a-3+b=0

=> 6-3a+b=0

=> -3a+b=-6 ---------------equation (2)

Now let's subtract equation (2) from (1).

I.e , 2a+b-(-3a+b)=-6-(-6)

So,2a+b+3a-b=-6+6

5a=0

Therefore,a=0

Then, from equation (1) we get:-

2a+b=-6

b=-6-2a -------------------equation (3)

Now by substituting the value of an in equation (3) we get:-

b=-6-2×0

Therefore,b=-6

Hence ,a=0 and b=-6

Explanation:

It is given in the above that the zeros of the polynomial are -2 and 5.

Now we have to find the number of polynomials that have zeros as - 2 and 5.

Here the polynomial is of the form

p(x) = ax² + bx + c

Now we know that,

Sum of zeros = – (coefficient of x) ÷ coefficient of x² = -b/a

-2 + 5 = -b/a

3 = -b/a

Therefore,b = -3 and a = 1

The product of the zeros = constant term ÷ coefficient of x² = c/a

c/a = (-2)(5)

Therefore, c = -10

Now let us substitute a, b, and c values in p (x)

x² - (3x) + (-10)

x² - 3x - 10

Therefore, the number of polynomials having zeros as -2 and 5 is more than 3.

Explanation:

Here we know that, if 𝛼, ꞵ and 𝛾 are the zeroes of a cubic polynomial ax³ + bx² + cx + d, then

𝛼 + ꞵ + 𝛾 = -b/a

𝛼ꞵ + ꞵ𝛾 + 𝛾𝛼 = c/a

𝛼ꞵ𝛾 = -d/a

given, 𝛼 = 0

Now, 𝛼ꞵ + ꞵ𝛾 + 𝛾𝛼 = (0)ꞵ + ꞵ𝛾 + 𝛾(0) = ꞵ𝛾

By using the property of polynomials,

ꞵ𝛾 = c/a

Therefore, the product of the other two roots is c/a.

Explanation:

Since -1 is a zero of x^3+ax^2+bx+c, we have

(−1)^3+a×(−1)^2+b×(−1)+c=0

⇒a−b+c−1=0

⇒c=1−a+b.

Also, a product of all zeros is given by

αβ×(−1)=−c

⇒αβ=c

⇒αβ=1−a+b.

Explanation:

We know that,

In a quadratic expression ax²+ bx + c

If a, b, c will have the same sign

then both zeroes of the

expression is negative

Compare given expression

x² + 99x + 127 with ax² + bx + c ,

a = 1 , b = 99 ,c = 127

All are positive signs,

Therefore,

The expressions both zeroes have

negative sign.

Correct answer B ) is correct.

Explanation:

The correct answer is A)

Product of zeroes = k

The sign is positive; it means both the zeroes have the same sign.

Sum of zeroes = -k

The sign is negative and both have the same sign hence the zeroes are both negative.

Explanation:

The correct answer is C)

It is given that the zeros of the quadratic polynomial ax^2 +bx+c,c≠0 are equal.

=> Value of the discriminant(D) has to be zero.

=>b^2 −4ac=0

=>b^2 =4ac

Since. L.H.S b^2  cannot be negative, thus, R.H.S. can also never be negative.

Therefore, a and c must be of the same sign.

Explanation:

(a) Let p(x) = x^2 + ax + b.

Put a = 0, then, p(x) = x^2 + b = 0

⇒ x^2 = -b

⇒ x = ± ±√-b

[∴b < 0]

Hence, if one of the zeroes of quadratic polynomial p(x) is the negative of the other, then it has no linear term i.e., a = O and the constant term is negative i.e., b< 0.

Alternate Method

Let f(x) = x^2 + ax +  b

and by given condition the zeroes area and – α.

The sum of the zeroes = α- α = a

=>a = 0

f(x) = x^2 + b, which cannot be linear,

and product of zeroes = α .(- α) = b

⇒ -α^2 = b

which is possible when b < 0.

Hence, it has no linear term and the constant term is negative.

Explanation:

For any quadratic polynomial ax² + bx + c, where a ≠0 the graph of the corresponding equation y = ax² + bx + c has one of the two shapes either open upwards like U or open downwards like inverted U (i.e. ∩) depending on whether a > 0 or a < 0.

In other words, the graph of a quadratic polynomial is a parabola.

A quadratic polynomial has less than or equal to two roots.

From the given options,

In Option D, the curve crosses the x-axis on three points.

This implies it has more than 3 roots, so it does not represent the quadratic polynomial.

Therefore, option D cannot be the graph of a quadratic polynomial.

Explanation:

no X -1 cannot be the remainder on the division of a polynomial p x by 2 X + 3 because the degree of the remainder cannot be the same as the degree of the divisor.

Explanation:

x^2 – 1 cannot be the quotient on a division of x^6 + 2x^3 + x – 1

 by a polynomial in x of degree 5. 

Justification:

 When a degree 6 polynomial is divided by a degree 5 polynomial, 

The quotient will be of degree 1. 

Assume that (x^2 – 1) divides the degree 6 polynomial and the quotient obtained is degree 5 polynomial (1) 

According to our assumption,

(degree 6 polynomial) = (x^2 – 1)(degree 5 polynomial) + r(x) [ Since, (a = bq + r)] 

= (degree 7 polynomial) + r(x) [ Since, (x^2 term × x^5 term = x^7 term)] 

= (degree 7 polynomial) 

From the above equation, it is clear that our assumption is contradicted. 

x^2 – 1 cannot be the quotient on division of x^6 + 2x^3 + x – 1 by a polynomial in x of degree 5 Hence Proved.

Explanation:

(i) False, if the zeroes of a quadratic polynomial ax^2+bx+c are both positive, then,

α+β=−b/a and α.β=c/a

where α and β are the zeroes of quadratic polynomials.

∴c<0a<0 and b>0 

or c>0a>0 and b<0

(ii) True, if the graph of a polynomial intersects the X-axis at only one point, then it cannot be a quadratic polynomial because a quadratic polynomial may touch the X-axis at exactly one point or intersects X-axis at exactly two points or do not touch then X-axis.

(iii) True, if the graph of a polynomial intersects the X-axis at exactly two points, then it may or may not be a quadratic polynomial. As, a polynomial of degree more than z is possible it intersects the X-axis at exactly two points when it has two real roots and other imaginary roots.

(iv) True, let α,β, and γ be the zeroes of the cubic polynomial and given that two of the zeroes have value 0.

Let β=γ=0

and f(x)=(x−α)(x−β)(x−γ)

=(x−α)(x−0)(x−0)

=x^3−ax^2

which does not have linear and constant terms.

(v) True, if f(x)=ax^3+bx^2+cx+d

 Then, for all negative roots, a,b,c, and d must have the same sign.

(vi) False, let α,β, and γ be the three zeroes of cubic polynomial 

x^3+ax^2−bx+c

Then, the product of zeroes 

=(−1)^3(Constant term/Coefficient of x^3)

⇒αβγ=−(+c)/1

⇒αβγ=−c ....(i)

Given that, all three zeroes are positive. So, the product of all three zeroes is also positive

i.e., αβγ>0

⇒−c>0 [from Eq.(i)]

⇒c<0

Now, the sum of the zeroes 

=α+β+γ=(−1)Coefficient ofx^2/Coefficient ofx^3

⇒α+β+γ=−a/1=−a

But α,β and γ are all positive.

Thus, its sum is also positive.

So, α+β+γ>0

⇒−a>0

⇒a<0

and sum of the product of two zeroes at a time =(−1)^2.Coefficient ofx/Coefficient ofx^3=−b/1

⇒αβ+βγ+γα=−b

∵αβ+βγ+αγ>0

⇒−b>0

⇒b<0

So, the cubic polynomial x^3+ax^2−bx+c has all three zeroes which are positive only when all constants a,b and c are negative.

(vii) False, let f(x)=kx^2+x+k

For equal roots. Its discriminant should be zero i.e., D=b^2−4ac=0

⇒k=±1/2

So, for two values of k, a given quadratic polynomial has equal zeroes.

Explanation:

Now we have given the polynomial: x^2 + 1/6(x) - 2 = 0

Simplifying it, we get:

               6x^2 + x - 12 = 0

So by splitting the middle term, we get:

               6x^2 - 8x + 9x - 12 = 0

Now combining the terms, we get:

               (6x^2 - 8x) + (9x - 12) = 0

               2x(3x - 4) + 3(3x + 4) = 0

               (3x - 4)(2x + 3) = 0

               x = 4/3 or x = -3/2

So verifying it, we get:

Sum of zeroes: 4/3 - 3/2 = -1/6

Product of zeroes: 4/3 x (-3/2) = -2

Hence verified.  So the zeroes of the polynomial are 4/3 and -3/2.

Explanation: 

The given equation is

4x²-3x-1 = 0

4x²-4x+x-1 = 0

4x(x-1)+1(x-1) = 0

(4x+1)(x-1) = 0

So, x₁ = 1, x₂ = -1/4

Now,

Sum of roots = -b/a = 3/4

Also, x₁+x₂ = 1-1/4 = 3/4

Product of zeroes = c/a = -1/4

Also, x₁*x₂ = 1(-1/4) = -1/4

Hence verified.

Explanation: 

Let 3x² + 4x - 4 = 0

On factoring,

= 3x² + 6x - 2x - 4= 3x(x + 2) - 2(x + 2)= (3x - 2)(x + 2)

Now, 3x - 2 = 0

3x = 2

x = 2/3

Also, x + 2 = 0

x = -2

Therefore, the zeros of the polynomial are 2/3 and -2.

We know that, if 𝛼 and ꞵ are the zeroes of a polynomial ax² + bx + c, then

The Sum of the roots is 𝛼 + ꞵ = -coefficient of x/coefficient of x² = -b/a

Product of the roots is 𝛼ꞵ = constant term/coefficient of x² = c/a

From the given polynomial,

coefficient of x = 4

Coefficient of x² = 3

Constant term = -4

The Sum of the roots:

LHS: 𝛼 + ꞵ

= 2/3 - 2= (2-6)/3= -4/3

RHS: -coefficient of x/coefficient of x²= -4/3

LHS = RHS

Product of the roots

LHS: 𝛼ꞵ= (2/3)(-2)= -4/3

RHS: constant term/coefficient of x²= -4/3

LHS = RHS.

Explanation:

5t^2 +12t+7=0

5t^2 +5t+7t+7=0

5t(t+1)+7(t+1)=0

(5t+7)(t+1)=0

⇒t= −7/5  and t=−1

∴ sum of roots is   −7/5−1  and t= −12/5

= −b/a 

product of roots = ( −7/5)(-1)=7/5

​= c/a

​Hence Verified.

Explanation: 

t^3 – 2t^2 – 15t 

Taking t common, we get, t ( t^2 -2t -15) 

Splitting the middle term of equation t^2 -2t -15, we get, 

t( t^2 -5t + 3t -15) Taking the common factors out, we get,

 t (t (t-5) +3(t-5) On grouping, we get, 

t (t+3)(t-5) So, 

the zeroes are, t=0 t+3=0 ⇒ t= -3 t -5=0 ⇒ t=5 

Therefore, zeroes are 0, 5 and -3 

Verification: Sum of the zeroes 

= – (coefficient of x2) ÷ coefficient of x3 α + β + γ 

= – b/a (0) + (- 3) + (5) = – (- 2)/1 = 2 = 2 

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x^3 αβ + βγ + αγ 

= c/a (0)(- 3) + (- 3) (5) + (0) (5) = – 15/1

 = – 15 = – 15

 Product of all the zeroes = – (constant term) ÷ coefficient of x3 αβγ 

= – d/a (0)(- 3)(5) = 0 

0 = 0.

Explanation:

2x^2 +(7/2)x +3/4

The equation can also be written as, 

8x^2+14x+3 

Splitting the middle term, we get, 

8x^2+12x+2x+3 

Taking the common factors out, we get, 4x (2x+3) +1(2x+3) 

On grouping, we get, (4x+1)(2x+3) 

So, the zeroes are, 

4x+1=0 ⇒ x = -1/4 2x+3=0 ⇒ x = -3/2 

Therefore, zeroes are -1/4 and -3/2 

Verification:

 Sum of the zeroes = – (coefficient of x) ÷ coefficient of x^2 

α + β = – b/a 

(- 3/2) + (- 1/4) = – (7)/4 

= – 7/4 = – 7/4

 Product of the zeroes = constant term ÷ coefficient of x^2 

α β = c/a 

(- 3/2)(- 1/4) = (3/4)/2 

3/8 = 3/8.

Explanation: 

P(X) = 4X²+5✓2X-3

=> 4X²+6✓2X-✓2X-3

=> 2✓2X(✓2X+3) -1(✓2X+3)

=> (✓2X+3) (2✓2X-1) = 0

=> (✓2X+3) = 0 OR (2✓2X-1) = 0

=> X = -3/✓2 OR X = 1/2✓2

-3/✓2 and 1/2✓2 are the two zeros of the given polynomial.

et Alpha = -3/✓2 and beta = 1/2✓2

Relationship between the zeros and Coefficient.

Sum of Zeros= (Alpha + Beta) = -3/✓2 + 1/2✓2 = -3×2✓2 + ✓2 = -6✓2+✓2/4 = -5✓2/4 = -( Coefficient of X/Coefficient of X².

And,

Product of zeros = (-3/✓2 × 1/2✓2) = -3/4 = Constant term/Coefficient of X².

Explanation: 

h(s) = 2s^2 – (1 + 2√2)s + √2 

 We put h(s) = 0  

⇒ 2s^2 – (1 + 2√2)s + √2 = 0 

 ⇒  2s^2 – 2√2s – s + √2 = 0  

⇒ 2s(s – √2) -1(s – √2) = 0  

⇒ (2s – 1)(s – √2) = 0 

 This gives us 2 zeros, for  x = √2 and x = 1/2  

Hence, the zeros of the quadratic equation are √3 and 1.  

Now, for verification  Sum of zeros = – coefficient of s / coefficient of s^2 

 √2 + 1/2 = – (-(1 + 2√2)) / 2  (2√2 + 1)/2 = (2√2 +1)/2 

 Product of roots = constant / coefficient of s^2  

1/2 x √2 = √2 / 2  √2 / 2 = √2 / 2  

Therefore, the relationship between zeros and their coefficients is verified.

Explanation:

f(v) = v^2 + 4√3v – 15  

We put f(v) = 0  ⇒ v^2 + 4√3v – 15 = 0  

⇒  v^2 + 5√3v – √3v – 15 = 0  

⇒ v(v + 5√3) – √3 (v + 5√3) = 0 

 ⇒ (v – √3)(v + 5√3) = 0 

 This gives us 2 zeros, for  v = √3 and v = -5√3  

Hence, the zeros of the quadratic equation are √3 and -5√3.  

Now, for verification  Sum of zeros = – coefficient of v / coefficient of v^2

  √3 + (-5√3) = – (4√3) / 1  -4√3 = -4√3  

Product of roots = constant / coefficient of v^2  

√3 x (-5√3) = (-15) / 1  -5 x 3 = -15  -15 = -15  

Therefore, the relationship between zeros and their coefficients is verified.

Explanation:

y^2 + 3√5/2y - 5 = 0 

2y^2 + 3√5y - 10 = 0 

2y^2 + (4√5y - √5y) - 10 = 0 

2y^2 + (4√5y - √5y) - 10 = 0 

2y(y + 2√5) - √5(y + 2√5)

 = 0 (y + 2√5)(2y - √5) = 0 

⇒ y = - 2√5, √5/2 

Verification: 

Sum of the zeroes = - (coefficient of x) ÷ coefficient of x^2 

α + β = - b/a (- 2√5) + (√5/2) = - (3√5)/2 

= - 3√5/2 = - 3√5/2

 Product of the zeroes = constant term ÷ coefficient of x^2

 α β = c/a (- 2√5)(√5/2)

 = - 5

 - 5 = - 5.

Explanation:

q(y) = 7y^2 – (11/3)y – 2/3

  We put q(y) = 0  

⇒ 7y^2 – (11/3)y – 2/3 = 0  

⇒  (21y^2 – 11y -2)/3 = 0  

⇒ 21y^2 – 11y – 2 = 0  

⇒ 21y^2 – 14y + 3y – 2 = 0  

⇒ 7y(3y – 2) – 1(3y + 2) = 0  

⇒ (3y – 2)(7y + 1) = 0  

This gives us 2 zeros, for 

 y = 2/3 and y = -1/7  

Hence, the zeros of the quadratic equation are 2/3 and -1/7.  

Now, for verification 

 The sum of zeros = – coefficient of y / coefficient of y^2  

2/3 + (-1/7) = – (-11/3) / 7 

 -11/21 = -11/21  

Product of roots = constant / coefficient of y^2

  2/3 x (-1/7) = (-2/3) / 7 

 – 2/21 = -2/21  

Therefore, the relationship between zeros and their coefficients is verified.

Explanation:

A quadratic polynomial in terms of the zeroes (α,β) is given by

x2 - (sum of the zeroes) x + (product of the zeroes)

i.e, f(x) = x2 -(α +β) x +αβ

Here, sum of the roots, α +β = √2

Product of the roots, αβ = 3/2

So, the quadratic polynomial can be written as x² - √2x - 3/2.

The polynomial can be rewritten as (1/2)[2x² - 2√2x - 3].

Let 2x² - 2√2x - 3 = 0

On factoring the polynomial,

2x² + √2x - 3√2x - 3 = 0

√2x(√2x + 1) - 3(√2x + 1) = 0

(√2x - 3)(√2x + 1) = 0

Now, √2x - 3 = 0

√2x = 3

x = 3/√2

Also, √2x + 1 = 0

√2x = -1

x = -1/√2

Therefore, the zeros of the polynomial are -1/√2 and 3/√2.

Explanation:

Let p ( x ) = x³ + 2x² + kx + 3 ,

If p( x ) is divided by ( x - 3 ) then the

the remainder is p ( 3 ),

But according to the problem given,

p( 3 ) = 21

3³ + 2( 3)² + k × 3 + 3 = 21

27 + 18 + 3k + 3 = 21

27 + 3k = 0

3k = - 27

k = ( -27 ) /3

k = - 9

Now,

x³ + 2x² + kx -18

= x³ + 2x² - 9x -18 = g ( x )

If x = -2

g ( x ) = ( -2 )³ + 2 ( -2 )² - 9 ( -2 ) - 18

= -8 + 8 +18 - 18

= 0

Therefore,

( x + 2 ) is a factor of g ( x )

g ( x ) = ( x + 2 ) ( x² - 9 )

= ( x + 2 ) ( x² - 3² )

= ( x + 2 ) ( x + 3 ) ( x - 3 )

Therefore ,

Zeros of g ( x ) is -2 , - 3 , 3

Explanation:

(i) -8/3, 4/3 

 A quadratic polynomial formed for the given sum and product of zeros is given by: 

 f(x) = x^2 + -(sum of zeros) x + (product of roots)  

Here, the sum of zeros is = -8/3 and product of zero= 4/3 

 Thus,  The required polynomial f(x) is,  

⇒ x^2 – (-8/3)x + (4/3)  

⇒ x^2 + 8/3x + (4/3)  

So, we put f(x) = 0 

 ⇒ x^2 + 8/3x + (4/3) = 0 

⇒ 3x^2 + 8x + 4 = 0  

⇒ 3x^2 + 6x + 2x + 4 = 0 

 ⇒ 3x(x + 2) + 2(x + 2) = 0  

⇒ (x + 2) (3x + 2) = 0  

⇒ (x + 2) = 0 and, or (3x + 2) = 0  

Therefore, the two zeros are -2 and -2/3. 

Explanation:

(ii) 21/8, 5/16 A quadratic polynomial formed for the given sum and product of zeros is given by:  

f(x) = x^2  - (sum of zeros) x + (product of roots) 

 Here, the sum of zeros is = 21/8 and product of zero = 5/16  

Thus, The required polynomial f(x) is,

  ⇒ x^2 – (21/8)x + (5/16)  

⇒ x^2 – 21/8x + 5/16  

So, We put f(x) = 0 

 ⇒ x^2 – 21/8x + 5/16 = 0  

⇒ 16x^2 – 42x + 5 = 0  

⇒ 16x^2 – 40x – 2x + 5 = 0 

⇒ 8x(2x – 5) – 1(2x – 5) = 0 

⇒ (2x – 5) (8x – 1) = 0  

⇒ (2x – 5) = 0 and, or (8x – 1) = 0 

 Therefore, the two zeros are 5/2 and 1/8.

Explanation:

 (iii) -2√3, -9 A quadratic polynomial formed for the given sum and product of zeros is given by:  

f(x) = x^2 - (sum of zeros) x + (product of roots)  

Here, the sum of zeros is = -2√3 and product of zero = -9  

Thus, The required polynomial f(x) is,  

⇒ x^2 – (-2√3)x + (-9) 

 ⇒ x^2 + 2√3x – 9  So, we put f(x) = 0  

⇒ x^2 + 2√3x – 9 = 0  

⇒ x^2 + 3√3x – √3x – 9 = 0  

⇒ x(x + 3√3) – √3(x + 3√3) = 0 

 ⇒ (x + 3√3) (x – √3) = 0  

⇒ (x + 3√3) = 0 and, or (x – √3) = 0  

Therefore, the two zeros are -3√3 and √3.

Explanation:

(iv) -3/2√5, -1/2  A quadratic polynomial formed for the given sum and product of zeros is given by:  

f(x) = x^2 - (sum of zeros) x + (product of roots) 

 Here, the sum of zeros is = -3/2√5 and product of zero = -1/2  

Thus,  The required polynomial f(x) is,  

⇒ x^2 – (-3/2√5)x + (-1/2)  

⇒ x^2 + 3/2√5x – 1/2  

So, we put f(x) = 0  

⇒ x^2 + 3/2√5x – 1/2 = 0  

⇒ 2√5x^2 + 3x – √5 = 0 

 ⇒ 2√5x^2 + 5x – 2x – √5 = 0

 ⇒ √5x(2x + √5) – 1(2x + √5) = 0  

⇒ (2x + √5) (√5x – 1) = 0  

⇒ (2x + √5) = 0 and, or (√5x – 1) = 0 

 Therefore, the two zeros are -√5/2 and 1/√5

Explanation:

Value of a = 5 & b = -3 or a = -1 & b = 3  and Zeroes are -1, 2, 5.

Step-by-step explanation:

Given: x³ - 6x² + 3x + 10

 Zeroes are of the form = a , a + b  , a + 2b

To find: Value of a & b and all zeroes of the polynomial.

Using the relationship of coefficient and Zeroes of the polynomial we get,

Sum of the roots = -coefficient of x²/ coefficient of x³

⇒ a + 2b + a + a + b = -(-6)/1                      

⇒ 3a + 3b = 6

⇒ 3 ( a + b ) = 6

⇒ a + b = 2 ..................(1)

Product of roots = -constant/coefficient of x³

⇒ ( a + 2b )( a + b ) a = -10/1

⇒ ( a + b + b )( a + b ) a = -10

⇒ ( 2 + b )( 2 ) a = -10   ( From equation (1) )

⇒ ( 2 + b ) 2a = -10

⇒ ( 2 + 2 - a ) 2a = -10

⇒ ( 4 - a ) 2a = -10

⇒ 4a - a² = -5

⇒ a² - 4a - 5 = 0

Now solving obtained quadratic equation we get,

a =5, -1                 

Now, Putting the value of an in the equation in Eqn (1)

We get,

When

a = 5   ⇒  5 + b = 2 ⇒ b = -3

a = -1   ⇒  -1 + b = 2 ⇒ b = 3

when a = 5 and b = -3,  Zeroes are 5 , ( 5 + (-3) ) = 2 , ( 5 + 2(-3) ) = -1

when a = -1 and b = 3 , Zeroes are -1 , ( -1 + 3 ) = 2 , ( -1 + 2(3) ) = 5

Therefore, the Value of a = 5 & b = -3 or a = -1 & b = 3, and Zeroes are -1, 2, 5.

Explanation:

Given, √2 is one of the zeros of the cubic polynomial.

Then, (x-√2) is one of the factors of the given polynomial p(x) = 6x³+√2x²-10x- 4√2.

Divide p(x) by x-√2

x-√2) 6x³+√2x²-10x-4√2 (6x² +7√2x

   6x³-6√2x²

(-)   (+)

  ----------------------------

7√2x² -10x-4√2

7√2x² -14x

 (-)       (+)

-------------------------

4x   - 4√2

4x   - 4√2

(-)    (+)

---------------------

0

6x³+√2x²-10x-4√2= (x-√2) (6x² +7√2x + 4)

= (x-√2) (6x² +4√2x + 3√2x + 4)

[By splitting middle term]

= (x-√2) [ 2x(3x+2√2) + √2(3x+2√2)

= (x-√2) (2x+√2)   (3x+2√2)

For zeroes of p(x), put p(x)= 0

(x-√2) (2x+√2)  (3x+2√2)= 0

x= √2 , x= -√2/2 ,x= -2√2/3

x= √2 , x= -1 /√2 ,x= -2√2/3

[ Rationalizing second zero]

Hence, the other two zeroes of p(x) are -1/√2 and -2√2/3.

Explanation:

Given, p(x) = 2x⁴ + x³ - 14 x² + 5x + 6.

g(x) = x² + 2x + k

We have to find the zeros of the polynomial.

The division algorithm states that given any polynomial p(x) and any non-zero

polynomial g(x), there are polynomials q(x) and r(x) such that

p(x) = g(x) q(x) + r(x), where r(x) = 0 or degree r(x) < degree g(x).

Let r(x) = 0

So, p(x) = g(x) q(x)

By using long division,

So, q(x) = 2x² - 3x - 8 - 2k

r(x) = (21+7k)x + (2k² + 8k + 6)

By comparing the coefficients of (21+7k)x and 2k² + 8k + 6

2k² + 8k + 6 = 0

2(k² + 4k + 3) = 0

k² + 3k + k + 3 = 0

k(k + 3) + (k + 3) = 0

(k + 1)(k + 3) = 0

Now, k + 1 = 0

k = -1

Also, k + 3 = 0

k = -3

So, k = -1, -3.

When k = -1,

21 + 7k = 0= 21 + 7(-1)= 21 - 7= 14

21 + 7k is not equal to zero.

So, k = -1 is neglected.

When k = -3,

21 + 7k = 0= 21 + 7(-3)= 21 - 21= 0

Therefore, the value of k is -3.

Now, g(x) = x² + 2x - 3

x² + 2x - 3 = 0

x² - x + 3x - 3 = 0

x(x - 1) + 3(x - 1) = 0

(x + 3)(x - 1) = 0

Now, x + 3 = 0

x = -3

Also, x - 1 = 0

x = 1

Now, q(x) = 2x² - 3x - 8 - 2k

= 2x² - 3x - 8 - 2(-3)= 2x² - 3x - 8 + 6= 2x² - 3x - 2

On factoring,

2x² - 3x - 2 = 0

2x² - 4x + x - 2 = 0

2x(x - 2) + (x - 2) = 0

(2x + 1)(x - 2) = 0

Now, 2x + 1 = 0

2x = -1

x = -1/2

Also, x - 2 = 0

x = 2

We know that g(x) and q(x) are the factors of p(x).

So, the zeros of g(x) and q(x) will be the zeros of p(x).

Therefore, the zeros of p(x) = -3, -1/2, 1 and 2.

Explanation:

given that x-√5 is a factor of the cubic polynomial x3-3√5x2+13x-3√5

x-√5 ) x3-3√5x2+13x-3√5 ( x2 -2√5x + 3

x3- √5x2 ( subtract )

-------------------------------

- 2√5x2+13x

- 2√5x2+10x ( subtract )

------------------------------

3x - 3√5

3x - 3√5 ( subtract )

------------------------

0

∴ The quotient is x2 -2√5x + 3 = 0

Using roots of the quadratic formula

a = 1, b = 2√5, c = 3

x = (-b ± √(b2 - 4ac) ) / 2a

x = (2√5 ± √((2√5)2 - 12) ) / 2

∴ the other zeros are x = √5 ± √2.

Explanation:

for zeroes of q(x) be also the zeroes of p(x), then q(x) should be the factor of p(x)

that is q(x) should wholly divide p(x) {with remainder 0}

p(x) = x⁵ - x⁴ -4x³ +3x² + 3x + b

q(x) = x³ +2x² +a

p(x) = q(x) . (x² -3x + 2)  + x²(-1-a) +x(3+3a) + b-2a

now remains to be zero implies

coefficient of x² = 0 ⇒ a = -1

coefficient of x = 0 ⇒ 3a = -3 ⇒ a = -1

and b-2a = 0 ⇒ b = -2

therefore p(x) = x⁵ - x⁴ -4x³ + 3x² - 2

and q(x) = x³+2x²-1

other zeroes of p(x) ;  solving (x² -3x +2 )=0

 x = 1 and x= 2.