# How do I factor this polynomial?

9x^2-12x+16

Sheet says I have to use difference of squares or perfect squares but everything I've tried hasn't worked.

The answer is not (3x-4^2) that solves out as

(3x-4) (3x-4)

9x2-12x-12x+16

9x-24x+16

not (3x-4)^2

### 7 Answers

- 1 month ago
Hmm...I'm assuming you want to solve this using algebra with the method of completing the square. Okay, here is what you do: 9X^2 - 12X + 16 = 0 =>

9X^2 -12X = -16 => X^2 - 4/3X = -16/9 => X^2 - 4/3X + 4/9 = -16/9 + 4/9 (where 4/9 is obtained by completing the square where you take the b term, 4/3, divide it by 2 giving 2/3, and then squaring that, giving 4/9) => (X-2/3)^2 = -12/9 => (X - 2/3)^2 = -4/3. Assuming you have learned complex numbers: X - 2/3 = (plus or minus) sqrt(-4/3) => X = 2/3 + sqrt(4/3) i or X = 2/3 - sqrt(4/3) i. With that, you can now factor the polynomial. All you need to do is some rearranging (by moving everything to the left side in your X equals equations and obtain the very ugly: (X - 2/3 - sqrt(4/3)i) (X - 2/3 + sqrt(4/3)i). You can make it a bit nicer by multiplying everything by 3 (ie X = 2/3 + sqrt(4/3)i => 3X = 2 + 3 sqrt(4/3) i => 3X = 2 + sqrt(4 * 9 / 3) i => 3X = 2 + 2 sqrt(3)i => 3X - 2 - 2sqrt(3)i = 0 and X = 2/3 - sqrt(4/3)i => 3X = 2 - 3sqrt(4/3)i => 3X = 2 - 2sqrt(3)i => 3X - 2 + 2sqrt(3)i). Thus, your answer would be: (3X - 2 - 2sqrt(3)i) (3X - 2 + 2sqrt(3)i). Notice that whenever a quadratic requires imaginary solutions, you will always get a complex number and its conjugate as solutions.

Now if you have not learned complex numbers yet, then the problem is a dud and the correct answer would be that there is no solution to the problem. As noted from comments down below, it is highly possible the problem had a typo: the equation should've been 9X^2 - 24X + 16. If that is the case, note that for the purposes of the answer, i = sqrt(-1). Observe that sqrt(a) * sqrt(b) = sqrt(a*b) for any real number a, b. Since we originally have sqrt(-4/3), we can rewrite it as sqrt(-1 * 4/3) = sqrt(-1) * sqrt(4/3) = sqrt(4/3)i. Even more, since i = sqrt(-1), i^2 = sqrt(-1) ^2 = -1. Hope this clears up any misconceptions.

- PinkgreenLv 71 month ago
This expression has no real factors.

9x^2-12x+16=

9(x^2-12x/9)+16=

9(x^2-4x/3)+16=

9[x^2-4x/3+(2/3)^2-(2/3)^2]+16=

{the method of completing the square}

[3(x-2/3)]^2-4+16=

[3(x-2/3)]^2+12=

[3(x-2/3)]^2-[sqr(12)i]^2=

{ i is the imaginary unit=sqr(-1); or i^2=-1}

[3(x-2/3)+2sqr(3)i][3(x-2/3)-2sqr(3)i]=

{a^2-b^2=(a+b)(a-b)}

[3x-2+2sqr(3)i][3x-2-2sqr(3)i]

- What do you think of the answers? You can sign in to give your opinion on the answer.
- Steve4PhysicsLv 71 month ago
You haven't stated the question exactly as given in the worksheet. So you've made it hard for us to answer.

Note (3x - 4)² = 9x² - 24x + 16, so this doesn't work.

9x² - 12x + 16 does *not* factor. So I think the question is about 'completing the square'.

We first want to find 'a' where (3x+a)² gives '-12' as the coefficient of x:

(3x+a)² = 9x² + 12ax + a²

therefore a = -1

(3x-1)² = 9x² - 12x + 1

If we add 15, we then get:

(3x-1)² +15 = 9x² - 12x + 1 + 15

(3x-1)² +15 = 9x² - 12x + 16 giving the final answer

9x² - 12x + 16 = (3x-1)² +15

- 1 month ago
maybe it means to complete the square? It’s a simple concept with a lot of helpful videos on youtube

- PuzzlingLv 71 month ago
There must be a typo in the worksheet. It seems they intended for you to work it as a perfect square:

The first term is a perfect square:

9x² = (3x)²

The last term is a perfect square:

16 = 4²

Then:

(a - b)² = a² - 2ab + b²

But if we have:

a = 3x

b = 4

Then the middle term is -2(3x)(4) --> -24x (not -12x).

It's a mistake.