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6x^2+5x-6
⇔
6x^2+5x+(- 6)
We have that a= 6, b=5, and c=- 6. There are now three steps we need to follow in order to rewrite the above expression.
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 &36 &- 1 + 36 &35 - 2 &18 &- 2 + 18 &16 - 3 &12 &- 3 + 12 &9 - 4 & 9 & - 4 + 9 &5 - 6 &6 &- 6 + 6 &0
Finally, we will factor the last expression obtained.
The GCF of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. In this case, the GCF is 2.
Do you notice that the binomial is a difference of two perfect squares? This can be factored using the difference of squares method. a^2 - b^2 ⇔ (a+b)(a-b) To do so, we first need to express each term as a perfect square.
| Expression | 4x^2-25 |
|---|---|
| Rewrite as Perfect Squares | (2x)^2 - 5^2 |
| Apply the Formula | (2x+5)(2x-5) |
Wait! Before we finish, remember that we factored out a GCF from the original expression. To fully complete the factored expression, let's reintroduce that GCF now. 2(2x+5)(2x-5)
The GCF of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. In this case, the GCF is 2x.
Split into factors
Factor out 2x
To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term. x^2+x- 56 In this case we have - 56. This is a negative number, so for the product of the constant terms in the factors to be negative, these constants must have the opposite sign — one positive and one negative.
| Factor Constants | Product of Constants |
|---|---|
| 1 and - 56 | - 56 |
| -1 and 56 | - 56 |
| 2 and - 28 | - 56 |
| -2 and 28 | - 56 |
| 4 and - 14 | - 56 |
| -4 and 14 | - 56 |
| 7 and - 8 | - 56 |
| - 7 and 8 | - 56 |
Next, let's consider the coefficient of the linear term. x^2+1x- 56 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 1.
| Factors | Sum of Factors |
|---|---|
| 1 and - 56 | - 55 |
| -1 and 56 | 55 |
| 2 and - 28 | - 26 |
| -2 and 28 | 26 |
| 4 and - 14 | - 10 |
| -4 and 14 | 10 |
| 7 and - 8 | - 1 |
| -7 and 8 | 1 |
We found the factors whose product is - 56 and whose sum is 1. x^2+1x- 56 ⇔ (x-7)(x+8) Wait! Before we finish, remember that we factored out a GCF from the original expression. To fully complete the factored expression, let's reintroduce that GCF now. 2x(x-7)(x+8)
| Is the first term a perfect square? | 9x^2=( 3x)^2 ✓ |
| Is the last term a perfect square? | 16= 4^2 ✓ |
| Is the middle term twice the product of 4 and 3x? | 24x=2* 4* 3x ✓ |
As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note there is a subtraction sign in the middle. 9x^2-24x+16 ⇔ ( 3x- 4)^2