Question

If (x - 1) and (x - 2) are factors of
$p {x}^{3} + 5 {x}^{2} + 6x + q$
find P and Q.
(A) p = -9/7 , q=16/7
(B) p=9/7 , q= 6/7
(C) p=-1/2 , q=16/7
(D) p=-2/7, q =1/8
(E) None of these​

Answer 1

$\huge\bf\underline{\underline{Question:}}$

If (x - 1) and (x - 2) are factors of$\sf{p {x}^{3} + 5 {x}^{2} + 6x + q}$find p and q.

$\huge\bf\underline{\underline{Solution:}}$

(x - 1) and (x - 2) are factors of p(x).

Therefore, by factor theorem, 1 and 2 are zeroes of p(x).

$\sf{\implies p(1)=p(2)=0}$

$\sf{\implies p(1)^3+5(1)^2+6(1)+q=p(2)^3+5(2)^2+6(2)+q}$

${\sf{\implies p+5+6+{\cancel q}=8p+20+12+{\cancel q}}}$

$\sf{\implies -7p=21}$

$\sf\red{\boxed{\therefore p = -3}}$

We know that p+11+q=0

Substituting the value of p

$\sf{\implies -3+11+q=0}$

${\sf\red{\boxed{\therefore q=-8}}}$

Therefore, (E)None of these is the correct option.

___________________________

@zaqwertyuioplm

Answer 2

HELLO, UR ANS:

option e) none of these