Find the value of a and b so that x²-4
is a factor of ax 4 + 2x3-3x2+bx-4.
Answers
Answer:
Let f(x) = ax4 +2x3 -3x2 +bx -4 and g(x) = x2 -4
We have g(x) = x2 − 4 = (x-4) (x+2)
Given g(x) is a factor of f(x).
(x-2) and (x+2) are factors of f(x)
From factor theorem,
If (x-2) and (x+2) are factor of f(x) then f(2) = 0 and f(-2) = 0 respectively
Answer:
Let us first factorize x
2
−4 as follows:
x
2
−4
=x
2
−2
2
=(x−2)(x+2)(∵a
2
−b
2
=(a−b)(a+b))
It is given that x
2
−4 is a factor of the polynomial f(x)=ax
4
+2x
3
−3x
2
+bx−4 that is (x−2)(x+2) are the factors of f(x)=ax
4
+2x
3
−3x
2
+bx−4 and therefore, x=−2 and x=2 are the zeroes of f(x).
Now, we substitute x=−2 and x=2 in f(x)=ax
4
+2x
3
−3x
2
+bx−4 as shown below:
f(−2)=a(−2)
4
+2(−2)
3
−3(−2)
2
+b(−2)−4
⇒0=16a−16−12−2b−4
⇒0=16a−2b−32
⇒2(8a−b−16)=0
⇒8a−b−16=0
⇒8a−b=16....(1)
f(2)=a(2)
4
+2(2)
3
−3(2)
2
+b(2)−4
⇒0=16a+16−12+2b−4
⇒0=16a+2b
⇒2(8a+b)=0
⇒8a+b=0....(2)
Adding equations 1 and 2:
(8a+8a)+(b−b)=16+0
⇒16a=16
⇒a=1
Substituting the value of a in equation 1, we get:
8a−b=16
⇒(8×1)−b=16
⇒8−b=16
⇒−b=16−8
⇒−b=8
⇒b=−8
Hence, a=1 and b=−8.
Step-by-step explanation:
I hope this helps you
kindly follow me