Quick math question(need to see if I worked question out right)

[Ricenchicken]

the question is:

Find K so that 4x+3 is a factor of: 20x^3+23x^2-10x+k

would I just replace x with -3/4

4x+3
4x=-3
x=-3/4

20(-3/4)^3+23(-3/4)^2-10(-3/4)+k
= 12+k
k=-12

answer is -12?

or am I just totally doing it wrong?

[Key_in_skillee]

You seem to have gotten the right answer. Here's how I did it:
The highest exponent in the polynomial is ^3 and the highest exponent in the factor is ^1, so the other factor is going to be a second degree polynomial.

(4x+3)(ax^2+bx+c) = 20x^3+23x^2-10x+k

Since there's only two things multiplied going into 20x^3 (4x*ax^2) we know a is 5.
(4x+3)(5x^2+bx+c) = 20x^3+23x^2-10x+k

now the two things that go into 23x^2 are 4x * bx and 3 * 5x^2, which is 15x^2. Subtract that from 23x^2 and you get 8x^2, which is what 4x*bx must equal. Therefore b must be 2.

(4x+3)(5x^2+2x+c) = 20x^3+23x^2-10x+k

the two things that go into -10x are 3 * 2x and 4x * c. 3*2x = 6x, so 4x * c must equal -16x for them to add up to -10x. Therefore, c is -4.

(4x+3)(5x^2+2x-4) = 20x^3+23x^2-10x+k

k ends up being the multiple of 3 and -4, so you get -12.

[Ricenchicken]

[QUOTE=Key_in_skillee;37839879]You seem to have gotten the right answer. Here's how I did it:
The highest exponent in the polynomial is ^3 and the highest exponent in the factor is ^1, so the other factor is going to be a second degree polynomial.

(4x+3)(ax^2+bx+c) = 20x^3+23x^2-10x+k

Since there's only two things multiplied going into 20x^3 (4x*ax^2) we know a is 5.
(4x+3)(5x^2+bx+c) = 20x^3+23x^2-10x+k

now the two things that go into 23x^2 are 4x * bx and 3 * 5x^2, which is 15x^2. Subtract that from 23x^2 and you get 8x^2, which is what 4x*bx must equal. Therefore b must be 2.

(4x+3)(5x^2+2x+c) = 20x^3+23x^2-10x+k

the two things that go into -10x are 3 * 2x and 4x * c. 3*2x = 6x, so 4x * c must equal -16x for them to add up to -10x. Therefore, c is -4.

(4x+3)(5x^2+2x-4) = 20x^3+23x^2-10x+k

k ends up being the multiple of 3 and -4, so you get -12.[/QUOTE]

Thank you! I started to try and factor I knew the first term was going to be squared but then I get jumbled up on my factoring. But I see it now.