Factor.

Question

Factor.

x, squared, minus, 12, x, plus, 27
x
2
−12x+27

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11 answers
(click or scroll down)

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Answer 1

Bot GPT 4-o mini answered
2 hours ago

To factor the quadratic expression \( x^2 - 12x + 27 \), we can look for two numbers that multiply to \( 27 \) (the constant term) and add to \( -12 \) (the coefficient of the \( x \) term).

The two numbers that meet these criteria are \( -3 \) and \( -9 \) since:

- \((-3) + (-9) = -12\)
- \((-3) \times (-9) = 27\)

Therefore, we can write the factorization as:

\[
x^2 - 12x + 27 = (x - 3)(x - 9)
\]

So, the factored form of \( x^2 - 12x + 27 \) is:

\[
(x - 3)(x - 9)
\]

Answer 2

anonymous answered
2 hours ago

Factor.

x, squared, plus, 4, x, minus, 21
x
2
+4x−21

Answer 3

Bot GPT 4-o mini answered
2 hours ago

To factor the quadratic expression \( x^2 + 4x - 21 \), we need to find two numbers that multiply to \( -21 \) (the constant term) and add up to \( 4 \) (the coefficient of the \( x \) term).

The two numbers that meet these criteria are \( 7 \) and \( -3 \) since:

- \( 7 + (-3) = 4 \)
- \( 7 \times (-3) = -21 \)

Using these numbers, we can factor the expression as follows:

\[
x^2 + 4x - 21 = (x + 7)(x - 3)
\]

So, the factored form of \( x^2 + 4x - 21 \) is:

\[
(x + 7)(x - 3)
\]

Answer 4

anonymous answered
2 hours ago

Factor the trinomial:

5, x, squared, plus, 7, x, plus, 2
5x
2
+7x+2

Answer 5

Bot GPT 4-o mini answered
2 hours ago

To factor the trinomial \( 5x^2 + 7x + 2 \), we can use the method of finding two numbers that multiply to the product of the first coefficient (\( 5 \)) and the last term (\( 2 \)), which is \( 5 \times 2 = 10 \), and add up to the middle coefficient (\( 7 \)).

The two numbers that satisfy these conditions are \( 5 \) and \( 2 \):

- \( 5 + 2 = 7 \)
- \( 5 \times 2 = 10 \)

Next, we rewrite the middle term \( 7x \) using these numbers:

\[
5x^2 + 5x + 2x + 2
\]

Now, we can group the terms:

\[
(5x^2 + 5x) + (2x + 2)
\]

Factoring out the common factors in each group, we get:

\[
5x(x + 1) + 2(x + 1)
\]

Now, we can factor out the common binomial factor \( (x + 1) \):

\[
(5x + 2)(x + 1)
\]

Thus, the factored form of \( 5x^2 + 7x + 2 \) is:

\[
(5x + 2)(x + 1)
\]

Answer 6

anonymous answered
2 hours ago

Factor the trinomial:

5, x, squared, plus, 32, x, plus, 12
5x
2
+32x+12

Answer 7

Bot GPT 4-o mini answered
2 hours ago

To factor the trinomial \( 5x^2 + 32x + 12 \), we need to find two numbers that multiply to \( 5 \times 12 = 60 \) (the product of the coefficient of \( x^2 \) and the constant term) and add up to \( 32 \) (the coefficient of the \( x \) term).

The two numbers that satisfy these conditions are \( 30 \) and \( 2 \):

- \( 30 + 2 = 32 \)
- \( 30 \times 2 = 60 \)

Now, we can rewrite the middle term \( 32x \) using these two numbers:

\[
5x^2 + 30x + 2x + 12
\]

Next, we group the terms:

\[
(5x^2 + 30x) + (2x + 12)
\]

Now, we can factor out the common factors in each group:

\[
5x(x + 6) + 2(x + 6)
\]

This gives us:

\[
(5x + 2)(x + 6)
\]

Thus, the factored form of \( 5x^2 + 32x + 12 \) is:

\[
(5x + 2)(x + 6)
\]

Answer 8

anonymous answered
2 hours ago

Solve the equation for all values of xx.

3, left parenthesis, 5, x, plus, 2, right parenthesis, left parenthesis, x, squared, minus, 81, right parenthesis, equals, 0
3(5x+2)(x
2
−81)=0

Answer 9

Bot GPT 4-o mini answered
2 hours ago

To solve the equation

\[
3(5x + 2)(x^2 - 81) = 0
\]

we can set each factor equal to zero.

1. The first factor:

\[
3 = 0
\]

This does not yield any solutions, as \(3\) is a non-zero constant.

2. The second factor:

\[
5x + 2 = 0
\]

Solving for \(x\):

\[
5x = -2 \implies x = -\frac{2}{5}
\]

3. The third factor:

\[
x^2 - 81 = 0
\]

This is a difference of squares, which can be factored as:

\[
(x - 9)(x + 9) = 0
\]

Setting each of these factors equal to zero gives us:

\[
x - 9 = 0 \implies x = 9
\]
\[
x + 9 = 0 \implies x = -9
\]

Combining all the solutions, we have:

\[
x = -\frac{2}{5}, \quad x = 9, \quad x = -9
\]

Thus, the complete solution set for the equation \( 3(5x + 2)(x^2 - 81) = 0 \) is:

\[
x = -\frac{2}{5}, \quad x = 9, \quad x = -9
\]

Answer 10

anonymous answered
1 hour ago

Solve the equation for all values of xx.

minus, 2, x, left parenthesis, x, plus, 8, right parenthesis, left parenthesis, x, squared, minus, 9, right parenthesis, equals, 0
−2x(x+8)(x
2
−9)=0

Answer 11

Bot GPT 4-o mini answered
1 hour ago

To solve the equation

\[
-2x(x + 8)(x^2 - 9) = 0,
\]

we can set each factor equal to zero:

1. The first factor:

\[
-2x = 0
\]

This gives:

\[
x = 0.
\]

2. The second factor:

\[
x + 8 = 0
\]

This gives:

\[
x = -8.
\]

3. The third factor:

\[
x^2 - 9 = 0.
\]

This can be factored as:

\[
(x - 3)(x + 3) = 0.
\]

Setting each of these factors equal to zero gives:

\[
x - 3 = 0 \implies x = 3,
\]
\[
x + 3 = 0 \implies x = -3.
\]

Now, combining all the solutions, we have:

\[
x = 0, \quad x = -8, \quad x = 3, \quad x = -3.
\]

Thus, the complete solution set for the equation

\[
-2x(x + 8)(x^2 - 9) = 0
\]

is:

\[
x = 0, \quad x = -8, \quad x = 3, \quad x = -3.
\]