# Advanced Factoring Techniques

Lesson Plans > Mathematics > Algebra > Factoring## Slide Show

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x^{2} - 6xy + 8y^{2}

x^{2}y^{4} - 4xy^{2} + 3

(x + y)^{2} +5(x + y) + 6

## Lesson Plan/Article

## Advanced Factoring Techniques

Many Algebra curricula have students practice primarily factoring polynomials in one variable, such as x^{2} - 7x + 12, which factors into (x - 3)(x - 4). When they do include multiple variables, the problem is often nothing more than factoring out common factors: a^{2}b + 2b^{2}a = ab(a+2b), or a perfect square factoring problem: x^{2} + 4xy + 4y^{2} = (x + 2)^{2}.

Students don't see many problems like this: x^{2} - 6xy + 8y^{2}. For my students, I explain that we're going to look at this as a quadratic in x, which means that y is actually part of the coefficient:

x^{2} - **6y**x + **8y ^{2}**

To complete this, we look for two expressions that add to -6y, and multiply to 8y^{2}. The expressions are -2y and -4y. Thus, this factors into (x - 2y)(x - 4y)

I think it's valuable for students to be able to do more advanced factoring problems with two variables that can be readily solved using substitutions. For example:

x^{2}y^{4} - 4xy^{2} + 3

If we let Z = xy^{2}, then this becomes z^{2} - 4z + 3 = (z - 3)(z - 1) = (xy^{2} -3)(xy^{2} - 1)

We can also use substitutions to handle problems with radicals or fractional exponents:

x + 4^{2}+ 4XY + 3Y

^{2}= (X + 3Y)(X + Y) = (

Similarly, if we see an expression like this:

(x + y)^{2} +5(x + y) + 6, we can make the substitution W = (x + y):

W^{2} + 5W + 6 = (W + 2)(W + 3) = (x + y + 2)(x + y + 3).

One of my students commented that this is like "breaking the rules." I asked what he meant, and he said, "Usually we use the distributive property." I agreed that this is what we *usually* do first, but it's not a rule that we have to. "So maybe *bending* the rules," he said. I replied, "No, more like using the rules creatively. The rules of algebraic manipulation are the same in every problem. The creativity comes in deciding which rules to apply, and when."

With that as background, here are some expressions students should be able to - with a little practice - factor.

## Handouts/Worksheets

## Factoring Worksheet

Factor the following expressions. Use substitution if needed, as shown below:**Example**: (x + y)

^{2}+5(x + y) + 6

**Solution**: Let W = (x + y). Then W

^{2}+ 5W + 6 = (W + 2)(W + 3) = (x + y + 2)(x + y + 3)

- a
^{2}b^{2}+ 8ab + 12

- a
^{2}- 7ab + 10b^{2}

- x
^{2}y^{4}- xy^{2}- 72

- x
^{1/2}- 5x^{1/4}- 14

- 2x + 9xy+9y

- 2x
^{5}- 11x^{3}+ 15x

- xy
^{3}- 3xyz +15z - 5y^{2}

- 3(m - 2n)
^{2}+ 7(m - 2n) + 4

- (a + b)
^{2}+ 4(a + b)(c + d) + 4(c + d)^{2}

- 5x
^{4}+ 9x^{2}y^{3}+ 4y^{6}

- x
^{6}+ 2x^{3}y^{3}z^{6}+ y^{6}z^{12}

- 3x
^{2/3}+ 16x^{1/3}+ 5

- k
^{2}- 2kx- 48x

- x + 5x
^{1/2}+ 6 + 2x^{2}+ 10x + 12

- 3x
^{7/3}+ 10x^{4/3}- 8x^{1/3}

- a
^{2}b^{4}cd - 9c^{3}d

- (a + b)
^{2}- (c + d)^{2}

- (2a - b)
^{2}- (b - 2c)^{2}

- a
^{2}+ 4ab + 4b^{2}- 4a - 8b + 3

- x + 2xy+ y + 4x+ 4y+ 4

## Factoring Worksheet: Answer Key

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