Algebra II - 5.5 Factoring

a^2-b^2=

(a+b)(a-b)

a^3+b^3=

(a+b)(a^2-ab+b^2)

a^2+2ab+b^2=

(a+b)^2

a^3-b^3=

(a-b)(a^2+ab+b^2)

a^2-2ab+b^2=

(a-b)^2

acx^2+(ad+bc)x+bd=

(ax+b)(x+d)

how to factor sums and differences of cubes

1) "un-cube" a^3 and b^3 in the given difference or sum of cubes (GCF if needed) 2) plug a and b into the appropriate formula depending on whether or not it's a difference or a sum

how to factor by grouping

1) factor out a GCF if there is one 2) factor the (four) terms into sets of binomials using parentheses 3) factor out a GCF from each set 4) the remaining binomials in each set should match; if not, check your factoring (if you cannot, and there is no GCF, it is fine) 5) write your final answer as the term inside the parentheses times the term you can make from the combination of the outside factors

how to factor a difference of squares in quadratic form

1) factor out the GCF if possible 2) identify the quadratic substitution for a and b 3) factor using a^2-b^2=(a+b)(a-b)

how to do general factoring in quadratic form

1) factor out the GCF if possible 2) identify the quadratic substitution u (always the variable+exponent in the second term) 3) substitute u in the equation for all the terms equivalent to the variable in the second term 4) completely factor the new equation that has u 5) substitute the original value of u back into the equation 6) factor this new equation completely

perfect square trinomials

a^2+2ab+b^2=(a+b)^2 a^2-2ab+b^2=(a-b)^2

binomial

a^2+b^2 if a and b share no common factors, then the sum of squares is not factorable with any real numbers

difference of squares

a^2-b^2=(a+b)(a-b)

factored sum of cubes

a^3+b^3=(a+b)(a^2-ab+b^2)

sum of cubes

a^3+b^3=(a+b)(a^2-ab+b^2)

difference of cubes

a^3-b^3=(a-b)(a^2+ab+b^2)

factored difference of cubes

a^3-b^3=(a-b)(a^2+ab+b^2)

general trinomials

acx^2+(ad+bc)x+bd=(ax+b)(x+d)

ways to factor with four or more terms

ax+bx+ay+by=x(a+b)(y+x)

how to factor with any number of terms

check for GCF

ways to factor with two terms

difference of squares sum of cubes difference of cubes

ways to factor with three terms

perfect square trinomials general trinomials

how to solve polynomials

use the zero product property to set each factor equal to zero and solve algebraically for real and imagined solutions when possible quadratic formula

quadratic formula

x = -b ± √(b² - 4ac)/2a

ax+bx+ay+by=

x(a+b)(y+x)