Let p(x)=x2+bx+c, where b and c are integers. If p(x) is a factor of both x4 + 6x2 + 25 and 3x4 + 4x² +28x + 5,
then what is the value of p(1) ?
(A) 0
(B) 1
(C) 2
(D) 4
Answers
Answer:
Step-by-step explanation:
p(x)=x2+bx+c (Given)
eq 1 - x4 + 6x² + 25
eq 2 - 3x4 + 4x² +28x + 5,
If(x) is a factor of P(x) and Q(x), then f(x) should be a factor of any linear combination of P(x) and Q(x). This is because, f(x) is individually a factor for both the functions.
Hence f(x) is factor of 3P(x) - Q(x) = 14x² - 28x + 70 = 14{x² - 2x + 5}
Now as f(x) is Quadratic equation, with co-eff od x² as 1, f(x) must be x² - 2x + 5.
= f(x) = (x-1)² + 4.
Thus, the minimum value of p1 will be 4
Step-by-step explanation:
ANSWER
P(x)=x
2
+bx+c is a factor of x
4
+6x
2
+25.
x
4
+6x
2
+25=x
4
+10x
2
−4x
2
+25=(x
4
+10x
2
+25)−4x
2
=(x
2
+5)
2
−(2x)
2
⇒x
4
+6x
2
+25=(x
2
+2x+5)(x
2
−2x+5)
One of them is a factor of 3x
4
+4x
2
+28x+5.
3x
4
+4x
2
+28x+5=3x
4
−6x
3
+15x
2
+6x
3
−11x
2
+28x+5
=3x
2
(x
2
−2x+5)+6x
3
−12x
2
+30x+x
2
−2x+5=3x
2
(x
2
−2x+5)+6x(x
2
−2x+5)+x
2
−2x+5
⇒3x
4
+4x
2
+28x+5=(x
2
−2x+5)(3x
2
+6x+1)
So, P(x)=x
2
−2x+5
P(x)=0⇒Δ=2
2
−4(5)=−16<0 has imaginary roots.
P(1)=1−2+5=4