Factor completely. 7z^(3)+14z^(2)-10 z-20

Identify Factors: Identify common factors in each term. We have the polynomial 7z^3 + 14z^2 - 10z - 20. We can look for common factors in pairs of terms. The first two terms, 7z^3 and 14z^2, have a common factor of 7z^2. The last two terms, -10z and -20, have a common factor of -10.Factor by Grouping: Factor by grouping. We group the terms with their common factors and factor them out: 7z^2(z + 2) - 10(z + 2). Now we can see that (z + 2) is a common factor.Factor Common Binomial: Factor out the common binomial factor. We can now factor out the common binomial factor (z + 2) from both groups: (7z^2 - 10)(z + 2).Check Further Factoring: Check for further factoring. The first group, 7z^2 - 10, does not factor further since 7 and 10 have no common factors and the expression is not a difference of squares or any other easily factorable form. The second group, z + 2, is already a binomial and cannot be factored further.

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Algebra 1

Factor polynomials

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Q. Factor completely.

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7z^{3}+14z^{2}-10z-20

Identify Factors: Identify common factors in each term. $\newline$ We have the polynomial $7z^3 + 14z^2 - 10z - 20$ . We can look for common factors in pairs of terms. $\newline$ The first two terms, $7z^3$ and $14z^2$ , have a common factor of $7z^2$ . $\newline$ The last two terms, $-10z$ and $-20$ , have a common factor of $-10$ .
Factor by Grouping: Factor by grouping. $\newline$ We group the terms with their common factors and factor them out: $\newline$ $7z^2(z + 2) - 10(z + 2)$ . $\newline$ Now we can see that $(z + 2)$ is a common factor.
Factor Common Binomial: Factor out the common binomial factor. $\newline$ We can now factor out the common binomial factor $(z + 2)$ from both groups: $\newline$ $(7z^2 - 10)(z + 2)$ .
Check Further Factoring: Check for further factoring. $\newline$ The first group, $7z^2 - 10$ , does not factor further since $7$ and $10$ have no common factors and the expression is not a difference of squares or any other easily factorable form. $\newline$ The second group, $z + 2$ , is already a binomial and cannot be factored further.