Answer:
A. 2
B. 2(5x -1)(x +3)
C. 2(5x -1)(x +3) = 2(5x² +14x -3) = 10x² +28x -6
Explanation:
You want the greatest common factor of 10x² +28x -6, its complete factorization, and a check of the factoring by multiplication.
A. GCD
There are several ways you can find the greatest common divisor (GCD) of the three coefficients in the given expression. Here, the simplest may be to factor those coefficients:
- 10 = 2 × 5
- 28 = 2 × 2 × 7
- 6 = 2 × 3
The one factor that appears on all the lists is 2, the greatest common divisor of these numbers.
The greatest common factor is 2.
B. Factorization
There are several ways you can factor the expression that is left after the GCD is factored out. In general, when factoring ax² +bx +c, you will be looking for factors of the product a·c that have a sum of b. Steps after that differ by factoring method. We'll show one method here.
expression = 10x² +28x -6 = 2(5x² +14x -3)
The reduced quadratic has a=5, b=14, c=-3, so we want factors of (5)(-3) = -15 that have a sum of 14.
-15 = 15(-1) = 5(-3) . . . . . ways to factor -15 with a positive sum of factors
The factor pairs have sums of 14 and 2. We're interested in the pair 15×(-1).
Using that sum to rewrite the x-term, we can write the reduced quadratic as ...
5x² +14x -3 = 5x² +(15x -x) -3
Now, we can group pairs of terms, and factor each group:
= (5x² +15x) +(-x -3) = 5x(x +3) -1(x +3)
The groups have a common factor, so we can write this as ...
= (5x -1)(x +3)
The complete factorization of the original expression is 2(5x -1)(x +3).
C. Check
We find it easiest to multiply the binomial terms first:
2(5x -1)(x +3)
= 2(5x² +15x -x -3) . . . . . "first", "outer", "inner", "last" term products
= 2(5x² +14x -3) . . . . . . . like terms combined
= 10x² +28x -6 . . . . distribute the 2; the original expression
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