Question

If f(x)= x³+4x² - 15-18 and a + 1 is a factor of f(x), then find all of the

zeros of f(x) algebraically?

Answer 1

Answer: x=-1 x=-6 x=3

Explanation:

`f(x)=x^3+4x^2-15x-18`

Let's isolate the multiplier (x+1) from the expression x³+4x²-15x-18 :

`x^3+4x^2-15x-18=\\\\x^3+x^2+3x^2-15x-18=\\\\x^2(x+1)+3(x^2-5x-6)=\\\\x^2(x+1)+3(x^2+x-6x-6)=\\\\x^2(x+1)+3(x(x+1)-6(x+1))=\\\\x^2(x+1)+3(x+1)(x-6)=\\\\(x+1)(x^2+3(x-6))=\\\\(x+1)(x^2+3x-18)`

Decompose the polynomial х²+3х-18 into factors:

`x^2+3x-18=\\\\x^2+6x-3x-18=\\\\x(x+6)-3(x+6)=\\\\(x+6)(x-3)`

`Thus,\\\\x^3+4x^2-15x-18=(x+1)(x+6)(x-3)\\\\`

`(x+1)(x+6)(x-3)=0\\\\x+1=0\\x+1-1=0-1\\x=-1\\\\x+6=0\\x+6-6=0-6\\x=-6\\\\x-3=0\\x-3+3=0+3\\x=3`

Answer 2

I believe you typed that wrong.

The third term should be -15X

And the 3 answers are 3 -6 -1

Explanation: