Question

Find the values of p and q so that (x+1) and (x-1) are the factors of polynomial x^4+px^3+2x^2-3x+q

Answer 1

$\textbf{ Hello Mate! }$

Gare up = If (x+1) and (x-1) neded to be made factors then value of x = - 1 and + 1 where fi al value of f(x) = 0.

Concept = We will make two equations, the add and then we will substitue the values.

Solution :-

$f(x) = {x}^{4} + p {x}^{3} + 2 {x}^{2} - 3x + q \\ f( - 1) = {( - 1)}^{4} + p( { - 1)}^{3} + 2 {( - 1)}^{2} - 3( - 1) + q \\ 0 = 1 - p + 2 + 3 + q \\ - 6 = - p + q \\ - 6 = - (p - q) \\ 6 = p - q \: ......(1)$

$f(1) = {(1)}^{4} + p( {1)}^{3} + 2 {(1)}^{2} - 3(1) + q \\ 0 = 1 + p + 2 - 3 + q \\ 0 = p + q \: ......(2)$

On adding (1) and (2) we get

6 = 2p

3 = p

Substituting value of p in eq (2)

0 = p + q

0 = 3 + q

- 3 = q

$\boxed{ \textsf{ \red{ Answer:\:p=3\:and\:q=\:-3 }}}$

Have great future ahead!

Answer 2

$\huge\underline\mathfrak{Answer:-}$

Let p(x) be the polynomial $\sf{x^4+px^3+2x^2-3x+q}$then,

$\sf{p(x)=x^4+px^3+2x^2-3x+q}$

$\sf{Since\:(x-1)\:is\:a\:factor\:of\:p(x)=}$

$\sf{x^4+px^3+2x^2-3x+q,then}$

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

$\sf{(1)^4+p(1)^3+2(1)^2-3(1)+q=0}$

$\sf{1+p+2-3+q=0}$

$\sf{p+q=0}$

$\sf{Again,\:(x+1)\:is\:a\:factor\:of\:p(x)=}$

$\sf{x^4+px^3+2x^2-3x+q,then}$

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

$\sf{(-1)^4+p(-1)^3+2(-1)^2-3(-1)+q=0}$

$\sf{1-p+2+3+q=0}$

$\sf{-p+q=-6}$

$\bold{\large{\boxed{\sf{\red{Adding\:(1)\:and\:(2)\:we\:get:-}}}}}$

$\sf{2q=0-6=-6}$

$\implies$$\sf{q=-3}$

But p+q=0

$\implies$$\sf{p=-q=-(-3)=3}$

$\bold{\large{\boxed{\sf{\pink{Thus,\:p=3\:and\:q=-3}}}}}$