Factor each completely. 6x^(2)+28 x+16

Find GCF: First, we look for a greatest common factor (GCF) that can be factored out from all the terms of the polynomial 6x^2 + 28x + 16. The GCF of 6, 28, and 16 is 2.Factor out GCF: We factor out the GCF from each term of the polynomial: 2(3x^2 + 14x + 8)Factor quadratic expression: Next, we need to factor the quadratic expression inside the parentheses. We look for two numbers that multiply to 3 * 8 (24) and add up to 14. The numbers that satisfy these conditions are 12 and 2 because 12 * 2 = 24 and 12 + 2 = 14.Write as binomials: We can now write the quadratic expression as a product of two binomials: 2(3x^2 + 12x + 2x + 8)Group terms to factor: We group the terms to factor by grouping: 2((3x^2 + 12x) + (2x + 8))Factor out common factors: Factor out the common factors from each group: 2(3x(x + 4) + 2(x + 4))Final factorization: We notice that (x + 4) is a common factor in both groups, so we factor it out: 2(3x + 2)(x + 4)

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Factor each completely. $\newline$ $6x^{2}+28x+16$

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Algebra 1

Evaluate integers raised to rational exponents

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Q. Factor each completely.

\newline

6x^{2}+28x+16

Find GCF: First, we look for a greatest common factor (GCF) that can be factored out from all the terms of the polynomial $6x^2 + 28x + 16$ . The GCF of $6$ , $28$ , and $16$ is $2$ .
Factor out GCF: We factor out the GCF from each term of the polynomial: $2(3x^2 + 14x + 8)$
Factor quadratic expression: Next, we need to factor the quadratic expression inside the parentheses. We look for two numbers that multiply to $3 \times 8$ ( $24$ ) and add up to $14$ . The numbers that satisfy these conditions are $12$ and $2$ because $12 \times 2 = 24$ and $12 + 2 = 14$ .
Write as binomials: We can now write the quadratic expression as a product of two binomials: $2(3x^2 + 12x + 2x + 8)$
Group terms to factor: We group the terms to factor by grouping: $2((3x^2 + 12x) + (2x + 8))$
Factor out common factors: Factor out the common factors from each group: $2(3x(x + 4) + 2(x + 4))$
Final factorization: We notice that $(x + 4)$ is a common factor in both groups, so we factor it out: $2(3x + 2)(x + 4)$