Question

Given the binomials (x - 3), (x - 1), (x + 2), and (x + 3), which one is a factor of f(x) = 2x3 + 3x2 - 2x - 3?

(x - 3)

(x - 1)

(x + 2)

(x + 3)

Answer 1

@SolomonZelman , could u take a look at this and just see how i need to set it up? Thank you.

Answer 2

if \(x-k\) for some number \(k\) is a factor of \(f\) that means \(f(x)=(x-k)\dots\) where the \(...\) is other stuff being multiplied by \(x-k\). Notice, then, that \(f(k)=(k-k)\dots=f(k)=0\times\dots=0\) i.e. if \(x-k\) is a factor then we know that \(f(k)=0\).

using this line of reasoning, consider \(x-3\) being a factor corresponds to \(f(3)=0\), \(x-1\) being a factor corresponds to \(f(1)=0\), \(x+2\) being a factor corresponds to \(f(-2)=0\), etc.

Answer 3

so here we have \(f(x)=2x^3 + 3x^2 - 2x - 3\). now, if \(x-3\) is a factor then we expect \(f(3)=0\). if we plug in \(3\) for \(x\), we see:$$f(3)=2(3)^3+3(3)^2-2(3)-3
\\f(3)=2\times27+3\times9-2\times3-3\\f(3)=54+27-6-3
\\f(3)=72\neq0$$
uhoh! clearly then we know that since \(f(3)\neq0\) then surely \(x-3\) is no factor of \(2x^3+3x^2-2x-3\).

Answer 4

and then you continue down the line; so, to check for \(x-1\), we plug in \(1\) for \(x\):$$f(1)=2(1)^3+3(1)^2-2(1)-3
\\f(1)=2+3-2-3\\f(1)=0$$and look there -- we have a factor!

Answer 5

Thank youu!!:) so x-1 is my answer?

Answer 6

now, the above approach is essentially the process of elimination, but you can answer this one in a more direct way as well. are you familiar with factoring by grouping? this particular problem lends itself to the technique quite well!

Observe: $$2x^3+3x^2-2x-3\\\underbrace{\color{blue}{2x^3+3x^2}}_\text{#1}\underbrace{\color{red}{-2x-3}}_\text{#2}\\$$now we take out common factors from #1 and #2 separately:$$\color{blue}{x^2(2x+3)}\color{red}{-1(2x+3)}$$and finally we pull out what both groups -- #1 and #2 -- have in common, in this case \(2x+3\):$$(2x+3)(\color{blue}{x^2}\color{red}{-1})$$

so far we have \(2x^3+3x^2-2x-3=(2x+3)(x^2-1)\). now, notice \(x^2-1=x^2-1^2\) is a difference of squares which we know factors like so:$$a^2-b^2=(a+b)(a-b)$$therefore $$(x^2-1)=(x+1)(x-1)$$ and so $$2x^3+3x^2-2x-3=(2x+3)(x^2-1)=(2x+3)(x+1)(x-1)$$

Answer 7

but yes, $x-1$ is your answer as seen above

Answer 8

Thank u!!