Question

find the value of a and b so that the polynomial p(x) and q(x) have (x2-x-12) as their hcf
where p(x)=(x2-5x+4)(x2+5x+a)
q(x)= (x2+5x+6)(x2-5x-2b)

Answer 1

so 3,-4 are the roots of eqn.

subs in p(x)

(9-15+4)(9+15+a)=0

(-2)(24+a)=0

-48-4a=0

-4a=48

a=-12

3 is subs in q(x)

(9+15+6)(9-15-2b)=0

-30(6+2b)=0

-180=60b

b=3

Answer 2

The values of a and b are 6, -2  

Given:

$p(x)=\left(x^{2}-5 x+4\right)\left(x^{2}+5 x+a\right)$ and $q(x)=\left(x^{2}+5x+6\right)\left(x^{2}-5x-2b\right)$have $x^{2} -x-12$ as their HCF

To find:

The values of a and b

Solution:

The polynomial p(x) and q(x) have HCF  $x^{2} -x-12$

⇒ $h(x)=(x-4) (x+3)$

∴ $4,-3$ are the factors for the $x^{2} -x-12$

Since, $p(x)=\left(x^{2}-5 x+4\right)\left(x^{2}+5 x+a\right)$

⇒p(x) = h(x)f(x), where, f(x) is a factor h(x)

Now, factorize the  $x^{2}-5 x+4$

$=(x-4)(x-1)$ .......(1)

⇒ $\left(x^{2}-5 x+4\right)\left(x^{2}+5 x+a\right)= (x-4)(x+3) f(x)$

Substitute equation (1) in above equation

⇒ $(x-4)(x-1)\left(x^{2}+5 x+a\right)=(x-4)(x+3)f(x)$

⇒ $(x-1)\left(x^{2}+5 x+a\right)=(x+3)f(x)$

Consider, $x=-3$

⇒$(-3-1)((-3)^{2} +5(-3)+a)=0$

⇒ $9-15+a=0$

⇒ $a=6$

Similarly, $q(x)=\left(x^{2}+5x+6\right)\left(x^{2}-5x+2b\right)$

⇒q(x) = h(x)g(x), where, g(x) is a factor h(x)

Now, factorize the $x^{2}+5x+6$

$=(x+3)(x+2)$........(2)

⇒ $\left(x^{2}+5 x+6\right)\left(x^{2}-5 x+2b\right)= (x-4)(x+3) g(x)$

Substitute equation (2) in above equation

⇒ $(x+3)(x+2)\left(x^{2}-5 x+2b\right)= (x-4)(x+3) g(x)$

⇒ $(x+2)\left(x^{2}-5 x+2b\right)= (x-4)g(x)$

Consider, $x=4$

⇒ $4^{2}-5 (4)+2b= 0$

⇒ $4+2b=0$

⇒ $b=-2$

Hence, the values of a and b are 6,-2

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