Warm Up 4-30
Simplify the Radical.
1) 5 27
2)
−50
3)
20
49
Solve each equation using the square root.
5) 𝑥 2 = 25
6) 𝑥 2 = −4
7) 3𝑥 2 − 27 = 0
4)
4
12
5
Complete the square.
8) 𝑥 2 − 16𝑥 − 36 = 0
Unit 7 Day 5: The
Zero-Product
Property
Essential Question: How can we use
factoring to solve polynomial equations?
To solve a polynomial equation in factored form,
we use the Zero Product Property.
If I’m thinking of 2 numbers that multiply to make
0, what can you tell me about those numbers?
One of them MUST be a zero (or both)!
The Zero Product Property states:If a • b = 0, then
a = 0 or b = 0
For example, the quadratic equation
15 = 0 in factored form is:
(2x – 3)(x + 5) = 0
2x - 3 = 0
x = 3/2
2
2x
+ 7x –
x+5=0
x = -5
These solutions also represent the x-intercepts of
a quadratic graph (parabola).
(3/2, 0) and (-5, 0)
Practice!
(x – 2)(x + 3) = 0
x – 2 = 0 or
+ 2+ 2
x=2
(x + 1)(x – 4) = 0
x+3=0
-3 -3
x + 1 = 0 or
-1 -1
x-4=0
+ 4+ 4
x = -3
x = -1
x=4
(2 , 0) and (-3 , 0)
(-1 , 0) and (4 , 0)
Practice!
(x +
2
5)
(x –
=0
2
8)
=0
x + 5 = 0 or
– 5– 5
x+5=0
– 5– 5
x – 8 = 0 or
+8+8
x–8=0
+8+8
x = -5
x = -5
x=8
x=8
(-5 , 0)
(8 , 0)
The entire point to all of this factoring that we
have been doing is so that you are able to find the
x-intercept solutions. If your equations are not in
factored form, factor them before you use the
Zero Product Property!
2
x
+ 3x - 4 = 0
(x - 1)(x + 4) = 0
x-1=0
(1 , 0)
x+4=0
(-4 , 0)
Practice!
x2 - 2x - 48 = 0
12 = 0 (x + 6)(x - 8) = 0
x + 6 = 0 or
– 6– 6
x–8=0
+ 8+ 8
x = -6
x=8
(-6 , 0) and (8 , 0)
2x2 + 14x +
2(x2 + 7x + 6) = 0
2(x + 6)(x + 1) = 0
x + 6 = 0 or
–6–6
x+1=0
–1–1
x = -6
x = -1
(-6 , 0) and (-1 , 0)
Practice!
3x2 + 17x + 10 = 0
(x + 5)(3x + 2) = 0
x + 5 = 0 or
- 5- 5
x = -5
3x + 2 = 0
-2-2
3x = -2
x = -2/3
(-5 , 0) and (-2/3 , 0)
8x2 + 8x - 6 = 0
2(2x + 3)(2x - 1) = 0
2x + 3 = 0or
- 3- 3
2x = -3
x = -3/2
2x - 1 = 0
+ 1+ 1
2x = 1
x = 1/2
(-3/2 , 0) and (1/2 , 0)
Sometimes, your equation will not be equal to 0.
This is a problem, because then we cannot use
the Zero Product Property to solve. To get around
this, we need to get everything on one side of the
equal sign so that our equation can equal 0.
2
x
+ 8x = -15
+ 15 + 15
2
x + 8x + 15 = 0
(x + 5)(x + 3) = 0
x+5=0
x+3=0
(-5 , 0)
(-3 , 0)
Practice!
x2 - 17x = -60
x2 – 3x = 10
– 10 – 10
+ 60 + 60
x2 - 17x + 60 = 0
x2 – 3x – 10 = 0
(x – 12)(x – 5) = 0
(x – 5)(x + 2) = 0
x – 12 = 0or
+ 12+ 12
x–5=0
+ 5+ 5
x – 5 = 0 or
+5+5
x+2=0
–2–2
x = 12
x=5
x=5
x = -2
(12 , 0) and (5 , 0)
(5 , 0) and (-2 , 0)
Summary
Respond to the essential question in the
summary portion of your notes.
Essential Question: How can we use
factoring to solve polynomial
equations?