Asked by Tiffany
12x^2+23x+10. Factor
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Answered by
GPT-4o mini
To factor the quadratic expression \(12x^2 + 23x + 10\), we can use the method of factoring by grouping.
1. Multiply the coefficient of \(x^2\) (which is 12) by the constant term (which is 10):
\[
12 \times 10 = 120.
\]
2. We need to find two numbers that multiply to 120 and add up to 23 (the coefficient of \(x\)). The two numbers that work are 15 and 8 because:
\[
15 \times 8 = 120 \quad \text{and} \quad 15 + 8 = 23.
\]
3. Rewrite the middle term (23x) using 15 and 8:
\[
12x^2 + 15x + 8x + 10.
\]
4. Now, group the terms:
\[
(12x^2 + 15x) + (8x + 10).
\]
5. Factor out the greatest common factor from each group:
\[
3x(4x + 5) + 2(4x + 5).
\]
6. Now, we can factor out the common binomial factor \((4x + 5)\):
\[
(4x + 5)(3x + 2).
\]
Thus, the factored form of \(12x^2 + 23x + 10\) is:
\[
(4x + 5)(3x + 2).
\]
1. Multiply the coefficient of \(x^2\) (which is 12) by the constant term (which is 10):
\[
12 \times 10 = 120.
\]
2. We need to find two numbers that multiply to 120 and add up to 23 (the coefficient of \(x\)). The two numbers that work are 15 and 8 because:
\[
15 \times 8 = 120 \quad \text{and} \quad 15 + 8 = 23.
\]
3. Rewrite the middle term (23x) using 15 and 8:
\[
12x^2 + 15x + 8x + 10.
\]
4. Now, group the terms:
\[
(12x^2 + 15x) + (8x + 10).
\]
5. Factor out the greatest common factor from each group:
\[
3x(4x + 5) + 2(4x + 5).
\]
6. Now, we can factor out the common binomial factor \((4x + 5)\):
\[
(4x + 5)(3x + 2).
\]
Thus, the factored form of \(12x^2 + 23x + 10\) is:
\[
(4x + 5)(3x + 2).
\]
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