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Math 097 - Quiz 7 Solutions
(1) Solve for p:
(5p − 2)(p + 7)(3p + 18) = 0
Since one side of the equation is zero, and the other side is a product, we can use the
zero product property: one of the factors on the left must be equal to 0.
5p − 2 = 0
5p = 2
or
or
p+7=0
p = −7
or
or
2
5
or
p = −7
or
p=
3p + 18 = 0
3p = −18
p = −6
So the solutions to this equation are p = 52 , −7, −6.
(2) Solve for x:
2x2 + 3x − 1 = x2 + 8x + 5
We’d like to use the zero product property again, but first we need to make one side
zero, and factor the other side (so it is a product instead of a sum).
2x2 + 3x − 1 = x2 + 8x + 5
−x2 − 8x − 5
− x2 − 8x − 5
x2 − 5x − 6 = 0
(x − 6)(x + 1) = 0
x−6=0
x=6
or
or
x+1=0
x = −1.
So the solutions are x = −1, 6.
(MORE ON BACK)
1
2
(3) Multiply and simplify the rational expressions:
2x + 12
x2 + 3x − 10
·
(x − 2)(x + 6)
2(x − 1)
Things aren’t quite factored all the way, so let’s finish factoring first. Then we can
multiply and simplify.
2x + 12
x2 + 3x − 10
·
(x − 2)(x + 6)
2(x − 1)
2(x + 6)
(x + 5)(x − 2)
=
·
(x − 2)(x + 6)
2(x − 1)
2(x + 6)(x + 5)(x − 2)
=
2(x − 2)(x + 6)(x − 1)
x+5
=
.
x−1
(4) Calculate the quotient of rational expressions:
(y + 3)2 (y − 7)3 (y + 3)2 (y − 1)
÷
(y − 1)2
y−7
Everything is already factored, so we can convert the division into multiplication, then
simplify.
(y + 3)2 (y − 7)3 (y + 3)2 (y − 1)
÷
(y − 1)2
y−7
2
3
(y + 3) (y − 7)
y−7
=
·
2
(y − 1)
(y + 3)2 (y − 1)
(y + 3)2 (y − 7)4
=
(y − 1)3 (y + 3)2
(y − 7)4
=
(y − 1)3