Products of Polynomials

Expanding

Polynomials can be expanded by multiplying each term. For example \((a + b)(c + d + e) = ab + ad + ae + bc + bd + be\)
If there are like terms, collect them: \((a + b)(a + c)(b + c) = (a + b)(ab + bc + ac + c^2) = a^2b + abc + a^2c + ac^2 + ab^2 + b^2c + abc + bc^2 = a^2b + a^2c + ab^2 + ac^2 + bc^2 + b^2c + 2abc\)

Factoring

There are many ways to factor and it really doesn’t matter which method is used:

Factor by Decomposition

\(6x^2 + 5x - 6 = 6x^2 + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = (2x + 3)(3x - 2)\)
Break the middle term down into two separate components then factor by grouping. Use common factoring to factor the non-simple trinomials after.

Factor by Decomposition Formula

\(6x^2 + 5x - 6 = \frac{(6x + ?)(6x + ?)}{6} = \frac{(ax + ?)(ax + ?)}{a}\)
2 numbers: product is -36 (6)(-6), sum is 5 (+9 and -4)
\(\frac{(6x + 9)(6x - 4)}{6} = \frac{3(2x + 3)2(3x - 2)}{6} = \frac{6(2x + 3)(3x - 2)}{6} = (2x + 3)(3x - 2)\)
Eliminate the numbers factored out from the numberator that cancel out the denominator.

Factor by Trial and Error

6 -2
1 +3

\((6)(3) + (1)(-2) = 18 - 2 \neq 5\)

6 +2
1 -3

\((6)(-3) + (1)(2) = -18 + 2 \neq 5\)

3 +2
2 -3

\((3)(-3) + (2)(2) = -9 + 4 \neq 5\)

3 -2
2 +3

\((3)(3) + (2)(-2) = 9 - 4 = 5 = (2x + 3)(3x - 2)\)

Factor by Box Method

2 numbers: product of \((6)(-6) = 36\) and sum of 5 which are +9 and -4

Place the two numbers in the corners

Factor each row and each column

= (2x + 3)(3x - 2)

Factor by 😱 Method

2 numbers with a product of \((6)(-6) = -36\) and sum of +5 (+9 and -4)

solve for both numbers:
\(9: \frac{a}{9}: \frac{6}{9} = \frac{2}{3} \to (2x + 3)\)
\(-4: \frac{a}{-4}: \frac{6}{-4} = \frac{3}{-2} \to (3x - 2)\)
= (2x + 3)(3x - 2)