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c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result 1 &30 &1 + 30 &31 2 &15 &2 + 15 &17 3 &10 &3 + 10 &13 5 & 5 & 5 + 6 &11
Split into factors
Factor out x
Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠ 1 and there are no common factors. To factor this expression, we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. x( 6x^2-31x+5 ) ⇔ x( 6x^2+(-31)x+5 ) We have that a= 6, b=-31, and c=5. 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 &30 &1 + 30 &31 -1 & -30 & -1 + ( -30) &-31
Split into factors
Factor out 3a
Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠ 1 and there are no common factors. To factor this expression, we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 3a( 2b^2+5b-7 ) ⇔ 3a( 2b^2+5b+(-7) ) We have that a= 2, b=5, and c=-7. 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 &14 &-1 + 14 &13 -2 & 7 & -2 + 7 &5
| Factor Constants | Product of Constants |
|---|---|
| 1 & -24 or -1 & 24 | -24 |
| 2 & -12 or -2 & 12 | -24 |
| 3 & -8 or -3 & 8 | -24 |
| 4 & -6 or -4 & 6 | -24 |
Next, let's consider the coefficient of the linear term. y^2+5y- 24 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 5.
| Factors | Sum of Factors |
|---|---|
| 1 and -24 | -23 |
| -1 and 24 | 23 |
| 2 and -12 | -10 |
| -2 and 12 | 10 |
| 3 and -8 | -5 |
| -3 and 8 | 5 |
We found the factors whose product is -24 and whose sum is 5. y^2+5y- 24 ⇔ (y-3)(y+8)