Question

I'm supposed to solve for x by factoring but I have no idea how to do that or where to start. I missed some information from previous lessons so I don't know how to even begin factoring anything.

Answer 1

`x = 3` or
`x = -5`

Explanation:

First, we need to expand the
`(x + 1)^2` term.

↓ rewriting the exponent as multiplication ...
`a^2 = a \cdot a`

`(x + 1)(x + 1)`

↓ rewriting using the distributive property ...
`a(b + c) = (a\cdot b) + (a\cdot c)`

`\left[\frac{}{}(x+1) \cdot x\frac{}{}\right] + \left[\frac{}{}(x+1) \cdot 1\frac{}{}\right]`

↓ simplifying using the distributive property

`(x^2 + x) + (x + 1)`

↓ combining like terms ...
`x + x = 2x`

`x^2 + 2x + 1`

↓ putting into the equation for
`(x + 1)^2`

`9(x^2 + 2x + 1) = 144`

Next, we can simplify the left side (once again) using the distributive property.

`(9\cdot x^2) + (9\cdot 2x) + (9 \cdot 1) = 144`

`9x^2 + 18x + 9 = 144`

Then, we can get all the terms on one side by subtracting 144 from both sides.

`9x^2 + 18x + 9 - 144 = 0`

`9x^2 + 18x - 135 = 0`

Finally, we can solve for x by factoring. To make this simpler, we can first factor out a 9 from each term on the left side.

`9(x^2 + 2x - 15) = 0`

We can think about this as "undistributing", or reversing the distributive property:

`(a \cdot b) + (a \cdot c) + (a \cdot d) = a(b + c + d)`

Next, we can factor the quadratic expression inside the parentheses using the rule:

`\text{if } x^2 + cx + d = (x + a)(x + b), \text{ then } a + b = c \text{ and } a \cdot b = d`

We can identify the following values for
`c` and
`d`:

• 
  `c=2`
• 
  `d=-15`

So, the following equations must be true:

`a + b = 2`,
`a \cdot b = -15`

We can solve for
`a` and
`b` by listing out the factor pairs of
`-15` and identifying the one whose factors add to
`2`.

• (-15, 1) →
  `-15 + 1 = -14` ❌
• (-5, 3) →
  `-5 + 3 = -2` ❌
• (-3, 5) →
  `-3 + 5 = 2` ✅

So, we can determine the following values for
`a` and
`b`:

• 
  `a=-3`
• 
  `b=5`

Hence, the factored form of the equation
`9(x^2 + 2x - 15) = 0` is:

`9(x - 3)(x + 5) = 0`

Now, we can solve for x using the zero product rule:

`\text{if } a \cdot b = 0, \text{ then } a = 0 \text{ or } b = 0`

Therefore, the two solutions for x are:

`x - 3 = 0` or
`x + 5 = 0`

`\boxed{x = 3}` or
`\boxed{x = -5}`