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Factor Constants | Product of Constants |
---|---|
1 and 7 | 7 |
-1 and - 7 | 7 |
Next, let's consider the coefficient of the linear term. x^2- 8x+7 For this term we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 8.
Factors | Sum of Factors |
---|---|
1 and 7 | 8 |
-1 and - 7 | - 8 |
We found the factors whose product is 7 and whose sum is - 8. x^2- 8x+7 ⇔ (x-1)(x-7)
Factor Constants | Product of Constants |
---|---|
1 and - 15 | - 15 |
-1 and 15 | - 15 |
3 and - 5 | - 15 |
-3 and 5 | - 15 |
Next, let's consider the coefficient of the linear term. y^2- 2y- 15 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 2.
Factors | Sum of Factors |
---|---|
1 and - 15 | - 14 |
-1 and 15 | 14 |
3 and - 5 | - 2 |
-3 and 5 | 2 |
We found the factors whose product is - 15 and whose sum is - 2. y^2- 2y- 15 ⇔ (y+3)(y-5)
The result of factoring out a GCF from the given expression is a binomial with a leading coefficient of 1. 7( x^2-9) 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 | x^2-9 |
---|---|
Rewrite as Perfect Squares | x^2 - 3^2 |
Apply the Formula | (x+3)(x-3) |
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. 7(x+3)(x-3)
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result 1 &24 &1+24 &25 2 &12 &2+12 &14 3 &8 &3+8 &11 4 & 6 & 4 + 6 &10