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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
The result of factoring out a GCF from the given expression is a binomial with a leading coefficient of 4. 2( 4x^2-25) Let's temporarily only focus on this binomial, and we will bring back the GCF after factoring.
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)
Split into factors
Factor out 2x
The result of factoring out a GCF from the given expression is a trinomial with a leading coefficient of 1. 2( x^2+x-56) Let's temporarily only focus on this trinomial, and we will bring back the GCF after factoring.
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 âś“ |
Is the last term a perfect square? | 16= 4^2 âś“ |
Is the middle term twice the product of 4 and 3x? | 24x=2* 4* 3x âś“ |
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