Question

if 2 is zero of both the polynomial 3x^2 +ax-4 and 2x^3+bx^2+x-2, then find the value of a+2b​

Answer 1

Answer:

If 2 is a zero of both the polynomials, then substitution of 2 in the respective polynomials will yield 0

P1(2) =8+2a=0

a = -4

P2(2)= 32+4b=0

b= -8

thus a+2b= -4+2(-8)

= -20

Answer: -20

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Answer 2

Let us first factorize x2-4 as follows:
X2-4
=x2-2x2

=(x−2)(x+2)

It is given that x2-4 is a factor of the polynomial f(x)=ax
4+2x3-3x2 + bx−4 that is (x−2)(x+2) are the factors of f(x)=ax

4+2x3-3x2 +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+2x3-3x2 +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.