For which values of $ a $ and $ b $, are the zeroes of $ q(x)=x^{3}+2 x^{2}+a $ also the zeroes of the polynomial $ p(x)=x^{5}-x^{4}-4 x^{3}+3 x^{2}+3 x+b $ ? Which zeroes of $ p(x) $ are not the zeroes of $ q(x) $ ?

Given:

\( q(x)=x^{3}+2 x^{2}+a \)

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

To find:

Here, we have to find the values of \( a \) and \( b \) such that the zeroes of \( q(x) \) are also the zeroes of the polynomial \( p(x) \).

Solution:

The zeroes of $q(x) = x^3 + 2x^2 + a$ are also the zeroes of the polynomial $p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b$.

This implies,

$q(x)$ is a factor of $p(x)$.

By using long division method, we get,

$x^3+2x^2+a$)$x^5 - x^4 - 4x^3 + 3x^2 + 3x + b$($x^2-3x+2$

                       $x^5+2x^4+ax^2$

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

                         $-3x^4-4x^3+(3-a)x^2+3x+b$

                       $-3x^4-6x^3-3ax$

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

                                 $2x^3+(3-a)x^2+(3+3a)x+b$

                                $2x^3+4x^2+2a$

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

                                 $-(1+a)x^2+(3+3a)x+(b-2a)$

If $(x^3 + 2x^2 + a)$ is a factor of $(x^5 - x^4 - 4x^3 + 3x^2 + 3x + b)$, then the remainder should be zero.

This implies,

$-(1 + a) x^2 + (3 + 3a) x + (b - 2a) = 0$

$= 0x^2 + 0x+0$

On comparing the coefficient of $x$, we get,

$3a + 3 = 0$

$3a = -3$

$a=-1$

$b - 2a = 0$

$b =2a$

$b = 2(-1)$

$= -2$

For $a = -1$ and $b = -2$, the zeroes of $q(x)$ are also the zeroes of the polynomial $p(x)$.

Therefore,

$q(x) = x^3 + 2x^2 -1$

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

Dividend $=$ Divisor $\times$ Quotient $+$ Remainder

$p(x) = (x^3 +2x^2 -1)(x^2 -3x + 2)+ 0$

$= (x^3 + 2x^2 -1)(x^2 -2x - x + 2)$

$= (x^3 + 2x^2 - 1) (x - 2) (x - 1)$

Hence, the zeroes of $p(x)$ are $l$ and $2$ which are not the zeroes of $q(x)$.

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Updated on: 10-Oct-2022

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