Question

Factorize for x^2+2x+5 to be a factor of x^4+px^2+q what must be the values of p and q​

Answer 1

Answer: p = 6 and q = 25

Step-by-step explanation:

$x^4+px^2+q$

= $x^2 (x^2+2x+5)-$$2x^3 - 5x^2$ $+px^2+q$

= $x^2(x^2+2x+5)-2x(x^2+2x+5)+4x^2+10x-5x^2+px^2+q$

= $x^2(x^2+2x+5)-2x(x^2+2x+5)+(p-1) x^2+10x+q$

= $x^2(x^2+2x+5)-2x(x^2+2x+5)+(p-1)(x^2+2x+5)-2(p-1).x-5(p-1)+10x+q$

= $(x^2+2x+5) (x^2–2x+p) +2(6-p).x+5(1-p)+q$

= Divisor × Q + R

Remainder = 0

2(6-p)x+(5–5p+q)= 0.

2(6-p)x+(5–5p+q)= 0.x + 0.

Equating the coefficient of x and constant term.

2(6-p) = 0 => p = 6 and

5–5p+q = 0.

5–5×6+q = 0.

q = 30–5 = 25

p = 6 and q = 25