QuestionAugust 26, 2025

15) 5x^2+15x-140 16) 10n^2+11n-8

15) 5x^2+15x-140 16) 10n^2+11n-8
15) 5x^2+15x-140
16)
10n^2+11n-8

Solution
4.1(266 votes)

Answer

5(x+7)(x-4); (5n-2)(2n+4) Explanation 1. Factor the first expression For 5x^{2}+15x-140, factor out the greatest common factor (GCF), which is 5: 5(x^{2}+3x-28). Then, factor the quadratic x^{2}+3x-28 by finding two numbers that multiply to -28 and add to 3. These numbers are 7 and -4. Thus, x^{2}+3x-28 = (x+7)(x-4). 2. Factor the second expression For 10n^{2}+11n-8, use the quadratic formula or trial and error to find factors. The expression can be factored as (5n-2)(2n+4) by finding pairs of factors that satisfy the equation.

Explanation

1. Factor the first expression<br /> For $5x^{2}+15x-140$, factor out the greatest common factor (GCF), which is 5: $5(x^{2}+3x-28)$. Then, factor the quadratic $x^{2}+3x-28$ by finding two numbers that multiply to -28 and add to 3. These numbers are 7 and -4. Thus, $x^{2}+3x-28 = (x+7)(x-4)$.<br /><br />2. Factor the second expression<br /> For $10n^{2}+11n-8$, use the quadratic formula or trial and error to find factors. The expression can be factored as $(5n-2)(2n+4)$ by finding pairs of factors that satisfy the equation.
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