Solving
Quadratic
Equations
Solving
Quadratic
Equations
8-6
8-6 by Factoring
by Factoring
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Holt
McDougal
Algebra 1Algebra
Algebra11
Holt
McDougal
Solving Quadratic Equations
8-6
by Factoring
Warm Up
Find each product.
1. (x + 2)(x + 7)
2. (x – 11)(x + 5)
x2 + 9x + 14
x2 – 6x – 55
3. (x – 10)2 x2 – 20x + 100
Factor each polynomial.
(x + 5)(x + 7)
4. x2 + 12x + 35
(x – 7)(x + 9)
5. x2 + 2x – 63
6. x2 – 10x + 16
(x – 2)(x – 8)
7. 2x2 – 16x + 32
2(x – 4)2
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Objective
Solve quadratic equations by factoring.
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
You have solved quadratic equations by graphing.
Another method used to solve quadratic equations is
to factor and use the Zero Product Property.
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Example 1A: Use the Zero Product Property
Use the Zero Product Property to solve the
equation. Check your answer.
(x – 7)(x + 2) = 0
x – 7 = 0 or x + 2 = 0
x = 7 or x = –2
The solutions are 7 and –2.
Holt McDougal Algebra 1
Use the Zero Product
Property.
Solve each equation.
Solving Quadratic Equations
8-6
by Factoring
Example 1A Continued
Use the Zero Product Property to solve the
equation. Check your answer.
Check (x – 7)(x + 2) = 0
(7 – 7)(7 + 2)
(0)(9)
0
0
0
0
Check (x – 7)(x + 2) = 0
(–2 – 7)(–2 + 2)
(–9)(0)
0
Holt McDougal Algebra 1
0
0
0
Substitute each solution
for x into the original
equation.
Solving Quadratic Equations
8-6
by Factoring
Example 1B: Use the Zero Product Property
Use the Zero Product Property to solve each
equation. Check your answer.
(x – 2)(x) = 0
(x)(x – 2) = 0
x = 0 or x – 2 = 0
x=2
The solutions are 0 and 2.
Check (x – 2)(x) = 0
(0 – 2)(0)
(–2)(0)
0
Holt McDougal Algebra 1
Use the Zero Product
Property.
Solve the second
equation.
(x – 2)(x) = 0
Substitute each (2 – 2)(2)
0
solution for x into
(0)(2)
0
the original
0
0
equation.
0
0
0
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 1a
Use the Zero Product Property to solve each
equation. Check your answer.
(x)(x + 4) = 0
x = 0 or x + 4 = 0
x = –4
Use the Zero Product
Property.
Solve the second equation.
The solutions are 0 and –4.
Check (x)(x + 4) = 0
(0)(0 + 4)
(0)(4)
0
Holt McDougal Algebra 1
(x)(x +4) = 0
Substitute each (–4)(–4 + 4)
0
solution for x into
(–4)(0)
0
the original
0
0
equation.
0
0
0
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 1b
Use the Zero Product Property to solve the
equation. Check your answer.
(x + 4)(x – 3) = 0
x + 4 = 0 or x – 3 = 0
x = –4 or
x=3
The solutions are –4 and 3.
Holt McDougal Algebra 1
Use the Zero Product
Property.
Solve each equation.
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 1b Continued
Use the Zero Product Property to solve the
equation. Check your answer.
(x + 4)(x – 3) = 0
Check (x + 4)(x – 3 ) = 0
(–4 + 4)(–4 –3) 0
(0)(–7)
0
0
0
Check (x + 4)(x – 3 ) = 0
(3 + 4)(3 –3)
(7)(0)
0
Holt McDougal Algebra 1
0
0
0
Substitute each
solution for x into
the original
equation.
Solving Quadratic Equations
8-6
by Factoring
If a quadratic equation is written in standard
form, ax2 + bx + c = 0, then to solve the
equation, you may need to factor before using the
Zero Product Property.
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Helpful Hint
To review factoring techniques, see lessons 8-3
through 8-5.
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Example 2A: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check
your answer.
x2 – 6x + 8 = 0
(x – 4)(x – 2) = 0
x – 4 = 0 or x – 2 = 0
Factor the trinomial.
Use the Zero Product
Property.
Solve each equation.
x = 4 or x = 2
The solutions are 4 and 2.
Check
Check
x2 – 6x + 8 = 0
x2 – 6x + 8 =
(4)2 – 6(4) + 8 0
(2)2 – 6(2) + 8
16 – 24 + 8 0
4 – 12 + 8
0 0
0
Holt McDougal Algebra 1
0
0
0
0
Solving Quadratic Equations
8-6
by Factoring
Example 2B: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check
your answer.
x2 + 4x = 21
x2 + 4x = 21
–21 –21
x2 + 4x – 21 = 0
The equation must be written in
standard form. So subtract
21 from both sides.
(x + 7)(x –3) = 0
Factor the trinomial.
x + 7 = 0 or x – 3 = 0
Use the Zero Product Property.
Solve each equation.
x = –7 or x = 3
The solutions are –7 and 3.
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Example 2B Continued
Solve the quadratic equation by factoring. Check
your answer.
x2 + 4x = 21
Check Graph the related quadratic function. The
zeros of the related function should be the same as
the solutions from factoring.
●
Holt McDougal Algebra 1
●
The graph of y = x2 + 4x – 21
shows that two zeros appear to
be –7 and 3, the same as the
solutions from factoring. 
Solving Quadratic Equations
8-6
by Factoring
Example 2C: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check
your answer.
x2 – 12x + 36 = 0
(x – 6)(x – 6) = 0
x – 6 = 0 or x – 6 = 0
x=6
or
x=6
Factor the trinomial.
Use the Zero Product Property.
Solve each equation.
Both factors result in the same solution, so there
is one solution, 6.
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Example 2C Continued
Solve the quadratic equation by factoring. Check
your answer.
x2 – 12x + 36 = 0
Check Graph the related quadratic function.
The graph of y = x2 – 12x + 36
shows that one zero appears to
be 6, the same as the solution
from factoring. 
●
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Example 2D: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check
your answer.
–2x2 = 20x + 50
The equation must be written in
–2x2 = 20x + 50
+2x2 +2x2
standard form. So add 2x2 to
0 = 2x2 + 20x + 50
both sides.
2x2 + 20x + 50 = 0
2(x2 + 10x + 25) = 0
2(x + 5)(x + 5) = 0
2≠0
or
x+5=0
x = –5
Holt McDougal Algebra 1
Factor out the GCF 2.
Factor the trinomial.
Use the Zero Product Property.
Solve the equation.
Solving Quadratic Equations
8-6
by Factoring
Example 2D Continued
Solve the quadratic equation by factoring. Check
your answer.
–2x2 = 20x + 50
Check
–2x2 = 20x + 50
–2(–5)2
–50
–50
Holt McDougal Algebra 1
20(–5) + 50
–100 + 50
–50 
Substitute –5 into the
original equation.
Solving Quadratic Equations
8-6
by Factoring
Helpful Hint
(x – 3)(x – 3) is a perfect square. Since both
factors are the same, you solve only one of
them.
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 2a
Solve the quadratic equation by factoring.
Check your answer.
x2 – 6x + 9 = 0
Factor the trinomial.
(x – 3)(x – 3) = 0
x – 3 = 0 or x – 3 = 0
Use the Zero Product Property.
Solve each equation.
x = 3 or x = 3
Both equations result in the same solution,
so there is one solution, 3.
Check
x2 – 6x + 9 = 0
Substitute 3 into the
original equation.
(3)2 – 6(3) + 9 0
9 – 18 + 9 0
0 0
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 2b
Solve the quadratic equation by factoring. Check
your answer.
x2 + 4x = 5
x2 + 4x = 5
–5 –5
x2 + 4x – 5 = 0
(x – 1)(x + 5) = 0
x – 1 = 0 or x + 5 = 0
x=1
or
x = –5
Write the equation in standard
form. Add – 5 to both sides.
Factor the trinomial.
Use the Zero Product Property.
Solve each equation.
The solutions are 1 and –5.
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 2b Continued
Solve the quadratic equation by factoring. Check
your answer.
x2 + 4x = 5
Check Graph the related quadratic function. The
zeros of the related function should be the same as
the solutions from factoring.
●
●
Holt McDougal Algebra 1
The graph of y = x2 + 4x – 5
shows that the two zeros
appear to be 1 and –5, the
same as the solutions from
factoring.
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 2c
Solve the quadratic equation by factoring. Check
your answer.
30x = –9x2 – 25
–9x2 – 30x – 25 = 0
–1(9x2 + 30x + 25) = 0
Write the equation in standard form.
–1(3x + 5)(3x + 5) = 0
Factor the trinomial.
–1 ≠ 0
Use the Zero Product Property.
– 1 cannot equal 0.
or
3x + 5 = 0
Factor out the GCF, –1.
Solve the remaining equation.
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 2c Continued
Solve the quadratic equation by factoring. Check
your answer.
30x = –9x2 – 25
Check Graph the related quadratic function. The
zeros of the related function should be the same
as the solutions from factoring.
●
Holt McDougal Algebra 1
The graph of y = –9x2 – 30x – 25
shows one zero and it appears to
be at
, the same as the
solutions from factoring.
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 2d
Solve the quadratic equation by factoring. Check
your answer.
3x2 – 4x + 1 = 0
(3x – 1)(x – 1) = 0
3x – 1 = 0 or x – 1 = 0
or x = 1
The solutions are
Holt McDougal Algebra 1
Factor the trinomial.
Use the Zero Product Property.
Solve each equation.
and x = 1.
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 2d Continued
Solve the quadratic equation by factoring. Check
your answer.
3x2 – 4x + 1 = 0
Check
3x2 – 4x + 1 = 0
3
–4
+1
0
Holt McDougal Algebra 1
0
0
Check
3x2 – 4x + 1 =
3(1)2 – 4(1) + 1
3–4+1
0
0
0
0
0
Solving Quadratic Equations
8-6
by Factoring
Example 3: Application
The height in feet of a diver above the water
can be modeled by h(t) = –16t2 + 8t + 8,
where t is time in seconds after the diver
jumps off a platform. Find the time it takes
for the diver to reach the water.
h = –16t2 + 8t + 8
0 = –8(2t2 – t – 1)
The diver reaches the water
when h = 0.
Factor out the GFC, –8.
0 = –8(2t + 1)(t – 1)
Factor the trinomial.
0=
–16t2
Holt McDougal Algebra 1
+ 8t + 8
Solving Quadratic Equations
8-6
by Factoring
Example 3 Continued
Use the Zero Product
–8 ≠ 0, 2t + 1 = 0 or t – 1= 0
Property.
2t = –1 or t = 1

It takes the diver 1 second
to reach the water.
Solve each equation.
Since time cannot be
negative,
does not
make sense in this
situation.
Check 0 = –16t2 + 8t + 8
0
0
0
–16(1)2 + 8(1) + 8 Substitute 1 into the
original equation.
–16 + 8 + 8
0
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 3
What if…? The equation for the height above
the water for another diver can be modeled
by h = –16t2 + 8t + 24. Find the time it takes
this diver to reach the water.
h = –16t2 + 8t + 24
0=
–16t2
+ 8t + 24
The diver reaches the water
when h = 0.
0 = –8(2t2 – t – 3)
Factor out the GFC, –8.
0 = –8(2t – 3)(t + 1)
Factor the trinomial.
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Check It Out! Example 3 Continued
–8 ≠ 0, 2t – 3 = 0 or t + 1= 0
2t = 3 or t = –1
t = 1.5
It takes the diver 1.5
seconds to reach the water.
Use the Zero Product
Property.
Solve each equation.
Since time cannot be
negative, –1 does not
make sense in this
situation.
Check 0 = –16t2 + 8t + 24
0
0
0
–16(1.5)2 + 8(1.5) + 24
Substitute 1 into the
–36 + 12 + 24
original equation.
0
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Lesson Quiz: Part I
Use the Zero Product Property to solve each
equation. Check your answers.
1. (x – 10)(x + 5) = 0 10, –5
2. (x + 5)(x) = 0
–5, 0
Solve each quadratic equation by factoring.
Check your answer.
3. x2 + 16x + 48 = 0 –4, –12
4. x2 – 11x = –24 3, 8
•
Holt McDougal Algebra 1
Solving Quadratic Equations
8-6
by Factoring
Lesson Quiz: Part II
5. 2x2 + 12x – 14 = 0 1, –7
6. x2 + 18x + 81 = 0
7. –4x2 = 16x + 16
–9
–2
8. The height of a rocket launched upward
from a 160 foot cliff is modeled by the
function h(t) = –16t2 + 48t + 160, where h
is height in feet and t is time in seconds.
Find the time it takes the rocket to reach the
ground at the bottom of the cliff.
5s
Holt McDougal Algebra 1