# Factor completely. 2x^2-9x+7

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Hi, Kristen,

There are several ways to factor these. Actually, most teachers use only one way, but many will tweek one key step various steps to make it more understandable for their students.

What I like to show is:

1) Determine what a, b, and c are.

The general expression for a quadratic is:

**ax^2 + bx + c**

We have:

**2x^2 - 9x + 7**

So, matching up the numbers to the letters, we have:

**a = 2, b = -9, c = 7**

2) Multiply a and c

**2*7 = 14**

3) This is the step that many teachers will tweek.

Determine two numbers that multiply to the result from "step 2" and add to b.

So, we need two numbers that multiply to 14. We would have:

2*7 and 1*14, both of them. But, also, we could have their negatives, also:

-2*-7 and -1*-14

So, we have four combinations:

**2 and 7**

**1 and 14**

**-2 and -7**

**-1 and -14**

But, only one combination add to -9, -2 + -7

So, those are our numbers we need:

**-2 and -7**

4) Next, rewrite the expression, breaking up the middle term into two separate terms. We have:

2x^2 - 9x + 7

So, we are breaking up the -9 with -2 and -7. We would have:

**2x^2 - 2x - 7x + 7**

5) Next, put parenthesis around the first 2 and last two terms. As in:

**(2x^2 - 2x) (- 7x + 7) *** Here, I like to think of the**

** middle part as:**

** (2x^2 -2x) + (-7x + 7)**

** with the addition sign in the**

** middle. I have seen it make a**

** huge difference for some**

** students. It's something like:**

** "x - 7" is also "x + -7"**

5) Factor each parenthesis. So, you can't get away from doing some factoring. Here:

**(2x^2 -2x) + (-7x + 7)**

We can take 2x out of the first parenthesis and -7 from the second parenthesis. That gives us:

**2x(x-1) + -7(x-1)**

Don't forget your "1" in the parenthesis. If done correct and if the expression can be factored, these parenthesis should be the exact same. Here, they are, so we are not only ok, but. . .

6) The parenthesis are one half of our answer. So:

**Answer**

2x**(x-1) **+ -7**(x-1) (x-1)**

What would "make up" our other parenthesis?

7) Everything outside the parenthesis is the other half of our answer. So, here, we have "2x" and "-7" outside the parenthesis. So, our other parenthesis is "2x - 7"

**Answer**

2x**(x-1) **+ -7**(x-1) (x-1)**(2x-7)

I hope this helps.

You need to factor completely the quadratic equation, such that:

`2x^2 - 9x + 7 = 2(x - x_1)(x - x_2)`

`x_1,x_2` represent the roots of equation `2x^2 - 9x + 7 = 0`

You may use quadratic formula to evaluate `x_1, x_2` , such that:

`x_(1,2) = (-b+-sqrt(b^2 - 4ac))/(2a)`

`a = 2, b = -9, c = 7`

`x_(1,2) = (9+-sqrt(81 - 56))/4`

`x_(1,2) = (9+-sqrt25)/4 => {(x_1 = (9+5)/4),(x_2 = (9-5)/4):}`

`{(x_1 = 7/2),(x_2 = 1):}`

`2x^2 - 9x + 7 = 2(x - 7/2)(x - 1)`

**Hence, evaluating the factored form of quadratic expression `2x^2 - 9x + 7` yields **`2x^2 - 9x + 7 = 2(x - 7/2)(x - 1).`