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Learn Exercise 2.4 with Free Lessons & Tips

Question 1:

Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Solution 1:

Let the zeroes be α, β,γ.

Given sum of the zeros=α+β+γ = 2

Sum of the product of it's zeros taken two at a time = αβ+βγ+γα = -7

Also, product of it's zeros= αβγ= -14

The cubic polynomial is of the form

x³ - (sum of the zeros) x² +(sum of the product of the zeros taken two at a time) - ( product of the zeros)=0

=> x³ - (α+β+γ)x² + (αβ+βγ+γα)x - (αβγ)=0

=> x³ -2x²-7x+14=0 which is the required polynomial.

Comments

Bhavani Ramesh

Question 2:

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:

(i) $2x^{3}+x^{2}-5x+2; \frac{1}{2}, 1, -2$ (ii) $x^{3}-4x^{2}+5x-2; 2,1,1$

Solution 2:

(i) On comparing the given polynomial with the polynomial ax³ + bx² + cx + d, we obtain a = 2, b = 1, c = -5, d = 2

$\therefore p(x)= 2x^{3}+x^{2}-5x+2$

$p(\frac{1}{2})= 2\times (\frac{1}{2})^{3}+\left ( \frac{1}{2} \right )^{2}-5\left ( \frac{1}{2}\right ) +2$

$\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2= 0$

Also, $p(1)= 2\times 1^{3}+1^{2}-5(1)+2$

2+1-5+2 = 0

Also, $p(-2)= 2(-2)^{3}+(-2)^{2}-5(-2)+2$

-16+4+10+2=0

Hence $\frac{1}{2}, 1 and -2$ are the zeroes.

α= $\frac{1}{2}$ β = 1 γ= -2

α+β+γ = $-\frac{1}{2}$ = $\frac{b}{a}$

αβ+βγ+γα = $-\frac{5}{2}= \frac{c}{a}$

αβγ = $-\frac{2}{2}= -\frac{d}{a}$

Thus, the relationship between the zeroes and the coefficients is verified.

(ii) On comparing the given polynomial with the polynomial ax^₃ + bx_² + cx + d, we obtain a = 1, b = -4, c = 5, d = -2.

$\therefore p(x)= x^{3}-4x^{2}+5x-2$

$p(2)= 2^{3}-4(2)^{2}+5(2)-2$

$8-16+10-2=0$

$p(1)= 1^{3}-4(1)^{2}+5(1)-2= 0$

Hence 2, 1 and 1 are the zeroes.

α= 2 β = 1 γ= 1

α+β+γ = $\frac{4}{1}=\frac{-b}{a}$

αβ+βγ+γα = $\frac{5}{1}= \frac{c}{a}$

αβγ = $\frac{2}{1}= -\frac{d}{a}$

Thus, the relationship between the zeroes and the coefficients is verified.

Comments

Tasneem

Question 3:

If the zeroes of the polynomial $x^{3}-3x^{2}+x+1$ are a – b, a, a + b, find a and b.

Solution 3:

Let p(x) = x³ - 3x² + x + 1
The zeroes of the polynomial p(x) are given as a - b, a, a + b.
Comparing the given polynomial with dx³ + ex² + fx + g, we can observe that
d = 1, e = -3, f = 1, g = 1

Sum of the zeroes = α+β+γ = $3=(a-b)+a+(a+b)$

$3a=3$

$a=1............(i)$

Product of the zeroes αβγ = $-1=(a-b)a(a+b)$

$a^{3}-ab^{2}=-1.........(ii)$

Substituting the value of a from (i) and (ii), we get,

$1^{3}-1b^{2}=-1$

$b^{2}=2$

$b=\pm \sqrt{2}$

Hence a=1 and $b=\pm \sqrt{2}$

Comments

Tasneem

Question 4:

If two zeroes of the polynomial $x^{4}-6x^{3}-26x^{2}+138x-35$ are $2\pm \sqrt{3}$ find other zeroes.

Solution 4:

The 2 zeroes which are given are $2+\sqrt{3} and 2-\sqrt{3}$

$\therefore \left \lfloor x-)(2+\sqrt{3}) \right \rfloor\left [ x-(2-\sqrt{3}) \right ]$

$(x-2)^{2}-(\sqrt{3})^{2}$

Therefore, x^₂ - 4x + 1 is a factor of the given polynomial.

For finding the remaining zeroes of the given polynomial, we will carry out the division

$x^{4}-6x^{3}-26x^{2}138x-35$

$(x^{2}-4x+1)(x^{2}-2x-35)$

$=(x-7)(x+5)$

Hence, 7 and -5 are also zeroes of the given polynomial.

Comments

Tasneem

Question 5:

If the polynomial $x^{4}-6x^{3}+16x^{2}-25x+10$ is divided by another polynomial $x^{2}-2x+10$ , the remainder comes out to be x + a, find k and a.

Solution 5:

By dividing $x^{4}$ - 6x³ + 16 x² - 25x + 10 by x² - 2x + k we get remainder

( -9+2k )x+10 - 8k + $k^{2}$

But ATQ, Remainder = x + a.

On comparing their coefficients,we have

2k - 9 = 1 ⇒ 2k = 10 ∴ k = 5

and, - ( 8 - k )k + 10 = a

⇒ a = - ( 8 - 5 )5 + 10

= - 3 × 5 + 10 = - 15 + 10 = - 5

∴ k = 5 & a = - 5

Comments

Manoj Kumar

All Exercises - Chapter 2 - Polynomials

Exercise 2.1

Exercise 2.2

Exercise 2.3

Exercise 2.4

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Learn Exercise 2.4 with Free Lessons & Tips

All Exercises - Chapter 2 - Polynomials

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