# If (x-2) and (x+2) are factors of 6x^3+ax^2+bx+16, determine a and b, and any remaining factors.?

So clearly this function has zeros at x=-2 and x=2, at least.

### 4 Answers

- micatkieLv 64 weeks agoFavourite answer
Let f(x) = 6x³ + ax² + bx + 16

(x - 2) is a factor of f(x). Then, f(2) = 0

6(2)³ + a(2)² + b(2) + 16 = 0

2a + b = -32 …… [1]

(x + 2) is a factor of f(x). Then, f(-2) = 0

6(-2)³ + a(-2)² + b(-2) + 16 = 0

2a - b = 16 …… [2]

[1] + [2]:

4a = -16

a = -4

[1] - [2]:

2b = -48

b = -24

f(x) = 6x³ - 4x² - 24x + 16

= 6x³ + 12x² - 16x² - 32x + 8x + 16

= (6x³ + 12x²) - (16x² + 32x) + (8x + 16)

= 6x²(x + 2) - 16x(x + 2) + 8(x + 2)

= (x + 2)(6x² - 16x + 8)

= 2(x + 2)(3x² - 8x + 4)

= 2(x + 2)(x - 2)(3x - 2)

The answers:

a = -4 and b = -24

Remaining factors: 2 and (3x - 2)

- VamanLv 74 weeks ago
6x^3+ax^2+bx+16=0. x=2 is aroot, Put it 6*8+4a+2b+16=0

4a+2b=-48-16=-64.

Now put x=-2

-48+4a-2b+16=0. This gives 4a-2b=32

2a+b=-32, 2a-b=16, Add the two. 4a=-16, a=-4,

For b use the first equation. -8+b=-32. b=-24. Please check the result.

- PhilipLv 64 weeks ago
Put f(x) = 6x^3+ax^2+bx+16;

Given : (x-2) & (x+2) are both factors of f(x);

Therefore (x-2)*(x+2) = (x^2-4) is a factor of f(x);

Then there exists a linear factor g(x) = (px + q) such that (x^2-4)(px +q) = f(x). Now

(x^2-4)(px+q) = px^3+qx^2-4px-4q = 6x^3+ax^2+bx+16. Equating coefficients of like

powers of x gives p = 6, q = a, -4p = b, -4q = 16. Then q = a = -4 and b = -24. Then

g(x) = (6x-4), f(x) = 6x^3 -4x^2 -24x +16 = (6x-4)(x^2-4) = 2(x+2)(3x-2)(x-2). I have

deliberately lined up the factors containing the zeros of f(x) in ascending order. which are -2, 2/3, 2.

- 4 weeks ago
(x - 2) * (x + 2) = x^2 - 4

(x^2 - 4) * (mx + n) = 6x^3 + ax^2 + bx + 16

mx^3 + nx^2 - 4mx - 4n = 6x^3 + ax^2 + bx + 16

mx^3 = 6x^3

m = 6

nx^2 = ax^2

n = a

-4mx = bx

-4m = b

-4 * 6 = b

-24 = b

-4n = 16

n = -4

a = -4

a = -4 , b = -24