Factor.

Question

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

Answer 1

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

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

Answer 3

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

Factor the trinomial:
5, x, squared, plus, 7, x, plus, 2
5x
2
+7x+2

Answer 5

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

Factor the trinomial:
5, x, squared, plus, 32, x, plus, 12
5x
2
+32x+12

Answer 7

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

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

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 $x$:

$$5x=-2⟹x=-\frac{2}{5}$$

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:

$$x-9=0⟹x=9$$ $$x+9=0⟹x=-9$$

Combining all the solutions, we have:

$$x=-\frac{2}{5},x=9,x=-9$$

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

$$x=-\frac{2}{5},x=9,x=-9$$

Answer 10

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

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:

$$x-3=0⟹x=3,$$ $$x+3=0⟹x=-3.$$

Now, combining all the solutions, we have:

$$x=0,x=-8,x=3,x=-3.$$

Thus, the complete solution set for the equation

$$-2x(x+8)(x^2-9)=0$$

is:

$$x=0,x=-8,x=3,x=-3.$$