Powers and Polynomials Reference
Factoring Polynomials
Factoring rewrites a polynomial as a product. Start with structure, not guessing: remove common factors first, count terms, then choose the pattern that matches what remains.
Solved Examples
Jump directly to a worked example by factoring pattern.
Fact Table
| What You See | Try | Reminder |
|---|---|---|
| Every term shares a factor. | GCF first. | Never skip this step. |
| Four terms. | Grouping. | Pair terms so both pairs reveal the same binomial. |
| Three terms. | Trinomial factoring. | For $x^2+bx+c$, find numbers that multiply to $c$ and add to $b$. |
| Two terms with subtraction and squares. | Difference of squares. | Creates conjugates. |
| Two terms with cubes. | Sum or difference of cubes. | Use SOAP for the signs in the second factor. |
| Need vertex form or a perfect square. | Complete the square. | Take half of $b$, then square it. |
Content Formulas
GCF
$$ab+ac=a(b+c)$$
Difference of Squares
$$a^2-b^2=(a-b)(a+b)$$
Trinomial Pattern
$$x^2+bx+c=(x+m)(x+n)$$ $$mn=c,\quad m+n=b$$
Perfect Square Trinomials
$$a^2+2ab+b^2=(a+b)^2$$ $$a^2-2ab+b^2=(a-b)^2$$
Cubes
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$ $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
Completing the Square
$$x^2+bx+\left(\frac b2\right)^2=\left(x+\frac b2\right)^2$$
SOAP is for cubes: Same sign in the binomial, Opposite sign next, Always Positive last.
Classic Examples
GCF First
Factor $6x^3-18x^2+12x$.
Solution Steps
- Find the greatest common factor.
- Factor completely and state the result.
$$6x^3-18x^2+12x=6x(x^2-3x+2)$$
$$6x^3-18x^2+12x=6x(x-1)(x-2)$$
Grouping
Factor $x^3+2x^2+3x+6$.
Solution Steps
- Group the four terms into pairs and factor each pair.
- Factor out the common binomial and state the result.
$$x^3+2x^2+3x+6=x^2(x+2)+3(x+2)$$
$$x^3+2x^2+3x+6=(x+2)(x^2+3)$$
Difference of Squares
Factor $16x^2-81$.
Solution Steps
- Rewrite the expression as a difference of squares.
- Apply the conjugate-factor pattern.
$$16x^2-81=(4x)^2-9^2$$
$$16x^2-81=(4x-9)(4x+9)$$
Basic Trinomial
Factor $x^2-3x+2$.
Solution Steps
- Find two numbers whose product is $2$.
- Confirm their sum is $-3$.
- Write the factored trinomial.
$$(-1)(-2)=2$$
$$-1+(-2)=-3$$
$$x^2-3x+2=(x-1)(x-2)$$
Leading Coefficient
Factor $2x^2+5x+2$.
Solution Steps
- Multiply the leading and constant coefficients.
- Find numbers with product $4$ and sum $5$.
- Split the middle term.
- Group and factor each pair.
- State the factored expression.
$$a c=2\cdot 2=4$$
$$4+1=5,\quad 4\cdot 1=4$$
$$2x^2+5x+2=2x^2+4x+x+2$$
$$2x^2+5x+2=2x(x+2)+1(x+2)$$
$$2x^2+5x+2=(2x+1)(x+2)$$
Perfect Square Trinomial
Factor $x^2+6x+9$.
Solution Steps
- Recognize the first and last terms as squares.
- Check that the middle term is twice their product.
$$x^2+6x+9=x^2+2(x)(3)+3^2$$
$$x^2+6x+9=(x+3)^2$$
Sum of Cubes
Factor $8x^3+1$.
Solution Steps
- Rewrite the expression as a sum of cubes.
- Apply the sum-of-cubes pattern using SOAP signs.
- Multiply out the second factor and state the result.
$$8x^3+1=(2x)^3+1^3$$
$$8x^3+1=(2x+1)((2x)^2-(2x)(1)+1^2)$$
$$8x^3+1=(2x+1)(4x^2-2x+1)$$
Difference of Cubes
Factor $27x^3-8$.
Solution Steps
- Rewrite the expression as a difference of cubes.
- Apply the difference-of-cubes pattern.
- Multiply out the second factor and state the result.
$$27x^3-8=(3x)^3-2^3$$
$$27x^3-8=(3x-2)((3x)^2+(3x)(2)+2^2)$$
$$27x^3-8=(3x-2)(9x^2+6x+4)$$
Complete the Square
Complete the square for $x^2-6x$.
Solution Steps
- Take half of the $x$ coefficient.
- Square that value.
- Add the perfect-square term and write the result.
$$\frac{-6}{2}=-3$$
$$(-3)^2=9$$
$$x^2-6x+9=(x-3)^2$$