Question

If x2 – 4 is a factor of 2x3 + ax2 + bx + 12, where a and b are constant. Then find the values of a and b.

Answer 1

GIVEN: p(x) = ax^4+2x^3–3x²+bx
g(x) = x²-4
Where p(x) & g(x) are polynomials in variable ‘x'
& here g(x) is a factor of p(x)
That means p(x) is exactly divisible by g(x)
Or p(x) is exactly divisible by factors of g(x)
Now, factors of g(x) = x²-4 = (x+2)(x-2)
So, p(x) is exactly divisible by (x+2) & (x-2)
That means, if p(x) is divided by (x+2) & then by (x-2) , the remainder has to be zero.
So now we find out the remainder in each case by remainder theorem:
If p(x) ÷(x+2) , the ramainder = p(-2)
ie, p(x)= ax^4+2x^3–3x²+bx is divided by (x+2)
Remainder= p(-2)= 16a - 16 -12 -2b =0………(1)
& if p(x)= ax^4+2x^3–3x²+bx is divided by (x-2)
Remainder= p(2)= 16a+16–12+2b=0……….(2)
eq(1) +eq(2)
=> 32a-24=0
=>32a =24
So a= 24/32 = 3/4………(3)
Now, eq(1) _ eq(2)
=> -32 -4b =0
=> 4b = -32
=> b= -8…………(4)
ANS a= 3 /4
b= —8

Answer 2

HEY MATE HERE IS THE ANSWER TO THE QUESTION....

▶️x2-4=0

x2=4, x=2

P(x)= 2x3 + ax2 + bx + 12

P(2)= 2*2*2*2* + a * 2*2 + b*2 +12

= 16 + 4a + 2b + 12=0
4a + 2b= - 28.
2a + b = - 14