Question

Find the value of p and q if x4+px3+2x2-3x+q is divisible by x2-1

Answer 1

it is given that polynomial x⁴ + px³ + 2x² - 3x + q is divisible by (x² - 1).

⇒x⁴ + px³ + 2x² - 3x + q is divisible by (x - 1)(x + 1)

⇒x⁴ + px³ + 2x² - 3x + q is divisible by both (x - 1) and (x + 1) .

from remainder theorem : when polynomial f(x) is divisible by (x - a) then, f(a) = 0

Let f(x) = x⁴ + px³ + 2x² - 3x + q

then, f(1) = 0, f(-1) = 0.

now, f(1) = 0

(1)⁴ + p(1)³ + 2(1)² - 3(1) + q = 0

⇒1 + p + 2 - 3 + q = 0

⇒p + q = 0.......(1)

and f(-1) = 0

(-1)⁴ + p(-1)³ + 2(-1)² - 3(-1) + q = 0

⇒1 - p + 2 + 3 + q = 0

⇒ p - q = 6 ......(2)

solve equations (1) and (2),

p = 3 and q = -3

Answer 2

$\bf{\underline{\underline{Answer}}}$

$\bf{Value\: of\: p = 3\: and\: q = -3.}$

$\bf{\underline{\underline{Step\: by\: step\: explanation}}}$

Given polynomial, $p(x)=x^4+px^3+2x^2-3x+q$

Factors of p(x) are x + 1 & x - 1

According to factor theorem which states that a linear polynomial of form x - a is factor of polynomial p(x) if p(a) = 0

So, we get

p(1) = 0

$1^4+p(1)^3+2(1)^2-3(1)+q=0$

1+p+2-3+q=0

p+q=0 .............(1)

p(-1) = 0

$(-1)^4+p(-1)^3+2(-1)^2-3(-1)+q=0$

1-p+2+3+q=0

q-p=-6 ....................(2)

Add eqn (1) & (2) we get

2q = - 6

⇒ q = -3

Put this in equation (1)

p + (-3) = 0

⇒ p = 3

Therefore, value of p = 3 and q = -3.