b If a second degree expression can be factored, we are able to rewrite the coefficient of the x-term as a sum of two numbers, a and b, whose product equals the expression's constant.
x^2+(a+b)x+ab
⇕
(x+a)(x+b)
From the expression, we can identify what this sum and product must be.
cccccc
x^2 & + & 12 & -7pt x & + & 36
x^2 & + & (a+b) & -7pt x & + & ab
Let's list all of the ways that 36 can be factored and investigate which pair of numbers, a and b, that sum to 12. Notice that because both the constant and the product are positive, both a and b must be positive.
c|c|c
integer & ab & a+b [0.3em]
36 & 1(36) & 37
36 & 2(18) & 20
36 & 3(12) & 15
36 & 4(9) & 13
36 & 6(6) & 12
When a and b are both 6, we have factored the quadratic.
x^2+12x+36 ⇔ (x+6)(x+6)
c To factor this quadratic, we will equate the expression with 0 and solve for the equations roots by using the Quadratic Formula.
Now we can factor the equation.
4x^2-4x-3 ⇔ 4(x+1/2)(x-3/2)
d Examining the quadratic, we see that both terms are perfect squares. Like in Part A, we can rewrite this expression as a difference of two squares and then factor it.
