Question

If 2 and -2 are the zeros of the polynomial p(x)= ax^4 + 2x^3- 3x^2 + bx - 4 find the value of a and b

Answer 1

Answer:

Therefore the value of a and b are 1 and -8 respectively.

Step-by-step explanation:

To solve this question we need to put 2 and -2 in place of x simultaneously and then obtain 2 equations from the polynomial.

We are given the polynomial $p(x)= ax^4 + 2x^3- 3x^2 + bx - 4$ if we put 2 firs we will get

$p(2)= a*2^4 + 2*2^3- 3*2^2 + 2b - 4 \\or, p(2)= 16a+16-12+2b-4\\or, p(2)= 16a+2b=0$

Now if we put -2 we will get

$p(-2)= a(-2)^4 + 2(-2)^3- 3(-2)^2 + b(-2) - 4\\\\or, p(-2)=16a-16-12-2b-4\\or, p(-2)=16a-2b-32 = 0$

Now adding

$p(2)+p(-2)=16a+2b+16a-2b-32=0+0\\or, 32a-32=0\\or, 32a=32\\a=1$

Similarly if we put a=1 in 16a+2b=0 we get

$16a+2b=0\\16+2b=0\\2b=-16\\b=-8$

Answer 2

Answer:

Value of a = 1 and b = -8

Step-by-step explanation:

2 and -2 are the zeroes of the polynomial p(x) = ax^4 + 2x^3 - 3x^2 +bx -4

Hence , p(2) must be 0

Now, p(2) = a ×2^4 + 2×2^3 - 3× 2^2 + b ×2 -4 = 16a + 16 -12 +2b -4 = 16a+2b

Hence , 16a+2b =0 ---------(1)

p(-2) is also 0

Now, p(-2)= a×(-2)^4 +2×(-2)^3 -3×(-2)^2 +b×(-2) -4 = 16a -16 -12 - 2b -4 = 16a - 2b -32

Hence, 16a - 2b -32 =0--------(2)

Now, from (1) + (2) ;

32a - 32 = 0 => a = 1

Now, by substituting a=1 in (1) we get ,

16+2b = 0 => 2b = -16 => b = -8

a=1 and b = -8