#### Andrew Rubin

##### New member
Hi everyone

I am doing a calculus course where my textbook is factorizing one step further than most online step-by-step calculators I have found without really explaining how. Now I am working on factorizing a cubic function, and I am stuck at one part. The book says:

$$\displaystyle 3x^2+8x-3=3\left(x-\frac{1}{3}\right)\left(x+3\right)$$

I get that you take out the GCF but I can't figure out the other steps behind this. Hopefully someone here can help.

#### willyengland

##### New member
1/3 and -3 are the zeros of the original function divided by 3.
Steps:
- set function = 0
- divide by 3
- zeros are 1/3 and -3 ==> (x-1/3)(x+3)

#### HallsofIvy

##### Elite Member
Having "taken out" 3, we have $$\displaystyle 3\left(x^2+ \frac{8}{3}- 1\right)$$. Typically, in "factoring" problems we have integer coefficients and seek factors with integer coefficients- there are simply too many possible factors if we allow fractions. Here we would need to find two numbers, a and b, such that a+ b= 8/3 and ab= -1. With b= 8/3- a, $$\displaystyle ab= 8/3a- a^2= -1$$ which gets us right back to the original equation. As willyengland suggests, first solve the equation without factoring, by completing the square or the quadratic formula, then form the factors.

#### lookagain

##### Senior Member
Having "taken out" 3, we have $$\displaystyle 3\left(x^2+ \frac{8}{3}- 1\right)$$.

Having factored out 3, we have $$\displaystyle \ 3\left(x^2+ \frac{8}{3}x - 1\right)$$.

#### JeffM

##### Elite Member
Hi everyone

I am doing a calculus course where my textbook is factorizing one step further than most online step-by-step calculators I have found without really explaining how. Now I am working on factorizing a cubic function, and I am stuck at one part. The book says:

$$\displaystyle 3x^2+8x-3=3\left(x-\frac{1}{3}\right)\left(x+3\right)$$

I get that you take out the GCF but I can't figure out the other steps behind this. Hopefully someone here can help.
If I don't "see" a factoring of a quadratic in about 15 seconds, I immediately resort to the quadratic formula.

$$\displaystyle d = 8^2 - 4(3)(-\ 3) = 64 + 36 = 100.$$

$$\displaystyle \therefore x = \dfrac{-\ 8 \pm \sqrt{100}}{2 * 3} = -\ 3 \text { or } \dfrac{1}{3}.$$

$$\displaystyle \therefore 3x^2 + 8x - 3 = 3(x + 3) \left ( x - \dfrac{1}{3} \right ).$$

DONE

EDIT: You do not say what the original cubic was so we cannot tell if there was a short cut available before you ever got to the quadratic.