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Solving Quadratic Equations by Factoring

Equations that can be written in the form Ax2 + Bx + C = 0

where a, b, and c are integers and a>0 are called quadratic equations . This form is called
the standard form.

The easiest way to solve quadratic equations is to factor (if possible) the polynomial on the left
side of the equation above . We then use the Zero- Product Theorem .

Zero- Product Theorem for any real numbers
a and b, if ab = 0, then a = 0 or b = 0.

Quadratic Equations :

1. x2 – 3x – 4 = 0 This quadratic equation is in standard form.
2. A2 = 6A – 9 This quadratic equation is not in standard form.
3. (x – 7)(x + 6) = –22 This is a quadratic equation not in standard form.
4. (x + 8)(x – 5) = 0 By the Zero-Product Theorem, the quadratic equation can be
solved.
5. x(x – 14) = 0 This quadratic equation is in factored form and equal to zero ;
it can be solved by using the Zero -Product Theorem.
To solve quadratic equations by factoring:

1. Put the equation into standard form.
2. Factor the polynomial .
3. Set each factor equal to zero.
4. Solve each of the first degree equations.
EXAMPLE 1: 2x2 – 9x – 35 =0
Solution: This is already in standard form, so we start by factoring.
Factor

Set each factor = 0
Solve each equation
The solution set is:
EXAMPLE 2: a2 = 6a – 9
Solution: We first put this into standard form.
The equation is not in
standard form
a2 = 6a – 9
Put into standard form a2 – 6a + 9 = 0
Factor
Set each factor = 0
Solve each equation
The solution set is: {3,3}
EXAMPLE 3: (t – 7)(t + 6) = –22
Solution: Multiple this out and put into standard form.
FOIL
Standard form
Factor
Set each factor = 0
Solve each equation t = 5 t = –4
The solution set is: {5,–4}
EXAMPLE 4: Solve x(x – 14) = 0
Solution The equation is in factored form and equal to zero . It can be
solved by using the Zero-Product Theorem.
Factor
Set each factor = 0
The solution set is: {0,14}
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