βˆ’

3
=
0
in the second row should be x+3=0
π‘₯
+
3
=Use the table to answer the question.

(x+3)(x+4)=0
(
π‘₯
+
3
)
(
π‘₯
+
4
)
=
0
xβˆ’3=0
π‘₯
βˆ’
3
=
0
x+4=0
π‘₯
+
4
=
0
x=3
π‘₯
=
3
x=βˆ’4
π‘₯
=
βˆ’
4
While using the Zero Product Property to find the values of x that make the quadratic equation (x+3)(x+4)
(
π‘₯
+
3
)
(
π‘₯
+
4
)
equals 0, Oliver completed the work provided in the table. Is Oliver’s work accurate?(1 point)
Responses

No, x+4=0
π‘₯
+
4
=
0
in the second row should be xβˆ’4=0
π‘₯
βˆ’
4
=
0
.
No, x plus 4 equals 0 in the second row should be x minus 4 equals 0 .

Yes, the work is accurate.
Yes, the work is accurate.

No, he should have a third column where he should set x=0
π‘₯
=
0
.
No, he should have a third column where he should set x equals 0 .

No, xβˆ’3=0
π‘₯
0
.

1 answer

Let's examine Oliver's work on the quadratic equation \((x + 3)(x + 4) = 0\).

According to the Zero Product Property, if the product of two factors is zero, then at least one of the factors must be zero. Therefore, Oliver correctly set each factor to zero, resulting in:

  1. \(x + 3 = 0\)
  2. \(x + 4 = 0\)

From these equations, he solved for \(x\):

  1. \(x + 3 = 0 \implies x = -3\)
  2. \(x + 4 = 0 \implies x = -4\)

However, in the provided response options, it seems like there is confusion about the equations:

  • The statement "x βˆ’ 3 = 0" is not a step in Oliver's work and is incorrect regarding the equation he started with.
  • The error about β€œx + 4 = 0 should be x βˆ’ 4 = 0” implies a misunderstanding of the provided factors.

Thus, the correct responses regarding his work should be:

  1. The equation x + 4 should be left as is, without changing it to x - 4.
  2. There is no requirement to set x = 0 as a separate column because the solutions arise directly from setting each factor to zero without any errors.

Therefore, the correct response is:

Yes, the work is accurate.

Oliver's work correctly applied the Zero Product Property without errors.