b Think of the process of factoring as multiplying two binomials in reverse.
C
c Rewrite the middle term as two terms.
D
d Rewrite the middle term as two terms.
A
a (2x+5)(x-1)
B
b (x+2)(x-3)
C
c (x+4)(3x+1)
D
d Not factorable, see solution.
Practice makes perfect
a 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^2+3x-5 ⇔ 2x^2+3x+(- 5)
We have that a= 2, b=3, and c=- 5. There are now three steps we need to follow in order to rewrite the above expression.
Find a c. Since we have that a= 2 and c=- 5, the value of a c is 2* (- 5)=- 10.
Find factors of a c. Since ac=- 10, which is negative, we need factors of a c to have opposite signs — one positive and one negative. Since their sum b=3, which is positive, the absolute value of the positive factor will need to be greater than the absolute value of the negative factor.
b 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- 6
In this case we have -6. 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 - 6
- 6
-1 and 6
- 6
2 and - 3
- 6
-2 and 3
- 6
Next, let's consider the coefficient of the linear term.
x^2-1x- 6
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 - 6
- 5
-1 and 6
5
2 and - 3
-1
-2 and 3
1
We found the factors whose product is - 6 and whose sum is -1.
x^2-1x- 6 ⇔ (x+2)(x-3)
c 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^2+13x+4
We have that a= 3, b=13, and c=4. There are now three steps we need to follow in order to rewrite the above expression.
Find a c. Since we have that a= 3 and c=4, the value of a c is 3* 4=12.
Find factors of a c. Since ac=12, which is positive, we need factors of a c to have the same sign — both positive or both negative — in order for the product to be positive. Since b=13, which is also positive, those factors will need to be positive so that their sum is positive.
d 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 try to rewrite the middle term, bx, as two terms. The coefficients of these two terms would have to be factors of ac whose sum must be b.
2x^2+5x+7
We have that a= 2, b=5, and c=7. There are now a few steps we need to follow in order to rewrite the above expression.
Find a c. Since we have that a= 2 and c=7, the value of a c is 2* 7=14.
Find factors of a c. Since ac=14, which is positive, we need factors of a c to have the same sign — both positive or both negative — in order for the product to be positive. Since b=5, which is also positive, those factors would need to be positive so that their sum is positive.
c|c|c|c
1^(st)Factor
&2^(nd)Factor
&Sum
&Result
1
&14
&1+14
&15
2
&7
&2+7
&9
We cannot find a pair of integers whose product is 14 and whose sum is 5. Therefore, we cannot use the general method for factoring trinomials. The given polynomial is not factorable.
