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Factoring binomials in Algebra 2 ... please help me

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debe32 | eNotes Newbie

Posted October 18, 2011 at 2:21 AM via web

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Factoring binomials in Algebra 2 ... please help me

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taylormath | High School Teacher | (Level 2) Adjunct Educator

Posted October 18, 2011 at 5:02 AM (Answer #1)

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Binomial factoring is done two different ways in Algebra 2 depending on the initial binomial.  The first step in factoring is to look for a common factor between the two terms. For instance, `2xy^3-10x` has a 2x in common with both terms. Factoring out the GCF (Greatest common factor) looks like:

`2xy^3-10x=2x(y^3-5)`

The factored answer can be checked using distributive property. Here are a couple more examples of factoring out the GCF.

a)  `6a^2b^2+18ab^3=6ab^2(a+3b)`

b)  `-12x^2z^4+8x^4z^3=4x^2z^3(-3z+2x^2)`

When a GCF does not exist between 2 terms, the binomial either forms a difference of two squares, sum/difference of two cubes, or is prime.

Difference of two squares:  `a^2-b^2=(a+b)(a-b)`

Example: `9x^2-49y^2=(3x+7y)(3x-7y)`

Sum of two cubes: `a^3+b^3=(a+b)(a^2-ab+b^2)`

Example: `x^3+64=(x+4)(x^2-4x+16)`

Difference of two cubes: `a^3-b^3=(a-b)(a^2+ab+b^2)`

Example: `27x^3-8y^3=(3x-2y)(9x^2+6xy+4y^2)`

The 6xy term above came from a=3x and b=2y, so ab=6xy.

For problems like: `9x^2+16y^2`

The answer is prime.  No GCF can be found between the two terms and no special relationship exists.

 

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