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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