If x² + 2x + 5 is a factor of x⁴ + Px²+ Q, then the value of P + Q is equal to
Answers
Step-by-step explanation:
Given: x^2 + 2x + 5 is a factor of x^4 + Px^2 + Q
x^2 + 2x + 5) x^4 + px^2 + q
x^4 + 2x^3 + 5x^2
----------------------------
-2x^3 + (p - 5)x^2 + q
-2x^3 - 4x^2 - 10x
---------------------------------------
(p - 1)x^2 + 10x + q
(p - 1)x^2 + (2p - 2)x + (5p - 5)
--------------------------------------------------------------
(10 - 2p + 2)x + (q-5p + 5)
(i)
= > 10 - 2p + 2 = 0
= > 12 - 2p = 0
= > p = 6.
(ii) q - 5p + 5 = 0
= > q - 5(6) + 5 = 0
= > q - 30 + 5 = 0
= > q - 25 = 0
= > q = 25.
Now,
p + q = 25 + 6
= 31.
Hence, the value of P + Q is 31.
Hope this helps!
Answer:
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 coeff. 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 , Answer.
and now the final: p+q = 25 + 6 = 31
HOPE IT HELPS!!
Step-by-step explanation: