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Factor Constants | Product of Constants |
---|---|
1 and - 30 | - 30 |
-1 and 30 | - 30 |
2 and - 15 | - 30 |
-2 and 15 | - 30 |
3 and - 10 | - 30 |
-3 and 10 | - 30 |
5 and - 6 | - 30 |
-5 and 6 | - 30 |
Next, let's consider the coefficient of the linear term. x^2+1x- 30 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 - 30 | - 29 |
-1 and 30 | 29 |
2 and - 15 | - 13 |
-2 and 15 | 13 |
3 and - 10 | - 7 |
-3 and 10 | 7 |
5 and - 6 | - 1 |
-5 and 6 | 1 |
We found the factors whose product is - 30 and whose sum is 1. x^2+1x- 30 ⇔ (x-5)(x+6)
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. - 3x^3+23x^2-14x ⇕ x ( - 3x^2+23x+(- 14) ) We have that a= - 3, b=23, and c=- 14. 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 &42 &1 + 42 &43 2 & 21 & 2 + 21 &23 3 &14 &3+14 &17 6 &7 &6 + 7 &13
We have that a= 2, b=- 5, and c=4. There are now a few 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 &- 8 &- 1 + (- 8) &- 9 - 2 &- 4 &- 2 + (- 4) &- 6 We cannot find a pair of integers whose product is 8 and whose sum is - 5. Therefore, we cannot use the general method for factoring trinomials. The given polynomial is not factorable.
Split into factors
Factor out 2x
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. 2x(3x^2+5x-12) ⇕ 2x( 3x^2+5x+(- 12)) We have that a= 3, b=5, and c=- 12. 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