Factoring Method of Quadratics

Learn To Use Factoring Method To Find A Parabola

Learning Goals

  • Learn the different forms of factoring
  • Learning how to use factoring to find the axis of symmetry
  • Learning how to use the factoring to find the x-intercepts

The Unit In Summary

  • In this Unit you will learn how to perform the factoring form, to solve word problems and to graph them.

Simple Factoring

  • The first way of solving the vertex form is through expanding and simplifying.
  1. You would have two terms in one bracket and another two terms in another bracket.
  2. Then to solve it you would Multiply the first term inside the first set of brackets, then you would multiply the first term in the first bracket by everything inside the second set of brackets.
  3. Then you would Multiply the second term in the first bracket by everything inside the second bracket.
  4. Finally you would simplify the equation you get from multiplying and that would be your answer.
  5. Example:
(2x+4)(4x+2)

=8x^2+4x+16x+8

=8x^2+20x+8




  • The next form is factoring binomials
  1. To solve this you would first need to find a term that can divide both terms.
  2. Then you would put the quotients in brackets and put the divisor on the outside of the bracket.
  3. Example: (4x+18)/2


=2(2x+9)



  • The next form is factoring simple Trinomials
The formula for this form is written as ax^2+bx+c, but because this is simple trinomials, there will be no "a" because a=1 so it would be written as just x^2+bx+c
  1. So in order to factor simple trinomials you would need to use a method called product and sum.
  2. To perform product and sum, you have to find two numbers that would sum up to "b" and the product of the two numbers would equal to "c".
  3. when you find those two numbers you would write them into the equation in place of "b". Then you would group the four terms into two groups then divide the terms based on the groups GCF.
  4. Then you would group the divisors together and then you would group together the quotients.
  5. Example- X^2+7x+10 2x5=10 2+5=7


=x^2+5x+2x+10

=x/(x^2+5x)+ 2/(2x+10)

=x(x+5)+2(x+5)

=(x+5)(x+2)




  • Factoring By grouping.
  • To Factor by grouping you would need to group like terms into their own groups.
  • Example-


ax+ay+2x+2y

=(ax+ay)+(2x+2y)

=a(x+y)+2(x+y)

=(x+y)+(a+2)

Complex Factoring

  • One of the methods that I use to solve Complex factoring like complex trinomials are Decomposition.
  1. Similar to simple tinomials, the only difference to the formula is that "a" does not equal 1.
  2. To perform decomposition first you multiply "a" by "c".
  3. Then you would use the sum and product method using the product of "a" and "c"
  4. Then you would take the two factors from the product of "a" and "c" and proceed to factoring the same way you did at this point in simple trinomials.
  5. Example-

8x^2+24x+18 *(8x18=144 12+12=24)

=(8x^2+12x) / 4x+(12x+18)6


=4x(2x+3) +6(2x+3)

=(2x+3)(4x+6)

Special cases

  • Factoring differences of squares
  1. To solve a difference of square binomial, you need to square root the squares and group them.
  2. Example-

25x^2-49

=(5x-7)(5x+7)



  • There is a Video on Perfect Square Trinomials Below by khan academy
Completing Perfect Square Trinomials

Using Factored Form In Graphing Situations

  • First to find the x-intercepts/ zeros.
  1. to find the x-intercept you would just put zero in place of y in each half of the factors
  2. Examples-

x^2+9x+18

=x^2+6x+3x+18

x 3

=x(x+6) +3(x+6)

=(x+6)(x+3)

0=(x+6)

-6=x

0=(x+3)

-3=x


Therefore, the zeros are (-3,0) and (-6,0)



  • Using the information from the previous example we can find the axis of symmetry.
  • The formula is x=r+s/2 then plug in the value of x to find the value of y
  • Example-


x=-3+-6

2

x=-4.5

y=-4.5^2+9(-4.5)+18

y=20.25-40.5+18

y=-2.25

Therefore, the axis of symmetry is (-4.5,-2.25)



  • To find the y intercept, you would plug 0 into the place of "x" and solve for y.
  • Example-


y=0^2+9(0)+18

y=1+18

y=19

Therefore the y intercept is (0,19)