Factorizing quadratic term

Andrew Rubin

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Joined
Jun 24, 2019
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16
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

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Jun 13, 2019
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29
1/3 and -3 are the zeros of the original function divided by 3.
Steps:
- set function = 0
- divide by 3
- solve quadratic equation
- zeros are 1/3 and -3 ==> (x-1/3)(x+3)
 

HallsofIvy

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Jan 27, 2012
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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

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Aug 22, 2010
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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

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Sep 14, 2012
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4,114
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.
 
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