```Solving a cubic function by
factoring: using the sum or
difference of two cubes.
By Diane Webb
What is a cube?
27
1
64
8
125
Factoring the sum or difference
of two cubes:
(a³+b³) a³ is a perfect cube since a*a*a = a³
b³ is a perfect cube since b*b*b = b³
Always remember that the factors of the sum or
difference of two cubes is always a (binomial) and a
(trinomial).
(a³+b³)=(binomial)(trinomial)
To find the two factors,
let’s do the following:
First the binomial: Take the cubed root of each monomial
within the problem.
l
Since, the cubed root of a³ = a and the cubed root of
b³ = b.
(a³+b³)=(a+b)(trinomial)
Now, the trinomial.
The first term of the trinomial is
the first term of the binomial squared.
The first term of the
binomial is “a” and a*a = a²
The second term of the trinomial is the opposite
of the product of the two terms of the binomial.
The product of “a” and “b” is “ab” and
(a³+b³)=(a+b)(a²-ab+__)
then the opposite tells you to
change the sign of the product.
The third term of the polynomial is the 2nd term
of the binomial squared.
(a³+b³)=(a+b)(a²+__+__)
(a³+b³)=(a+b)(a²-ab+b²)
The second term of the binomial is
“b” and b*b=b²
Factor (x³-8)
Binomial factor is (x-2)
Trinomial factor is (x²+2x+4)
Remember that the trinomial is not factorable.
Factored form: (x³-8)=(x-2)(x²+2x+4)
Is it possible to check our answers?
Remember that you may check using either the Remainder
Theorem or division.
Remainder theorem: If P(x)=x³-8 and the factor is (x-2),
Then P(2)=(2)³-8
=8–8
=0
Since the remainder is 0, then x-2 is a factor.
Synthetic division:
2 1 0 0 -8
2 4 8
1 2 4 0
This tells you two things: 1. (X-2) is a factor since the
remainder is 0. 2. The quotient is x²+2x+4 which is
the trinomial factor of the cubic polynomial.
First of all, you can not forget that the
GCF must be factored out of the cubic
function. SO, what is the GCF of
2x³ and 2?
2 is the GCF. Now, factor
the two first: 2(x³+1)
Look at the binomial. Is it a difference
or the sum of two cubes. If yes, factor the
expression.
2x³+2 = 2(x³+1) = 2(x+1)(x²-x+1)
Now that we can factor the
sum and difference of two
cubes, let us solve them.
Remember to solve a cubic equation, we
need to use our factors. Set your
factors equal to 0. Then solve for x.
• Go back to the the previous problem:
• 2x³+2=0
•
•
•
•
•
•
Factored form: 2(x+1)(x²-x+1)=0
Set the factors equal to 0.
2=0 (x+1)=0 (x²-x+1)=0
20 so it is not part of our solution.
X+1=0 so x = -1.
What about x²-x+1=0? Remember we talked
about the fact that it is not factorable. How
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