Katelynn Cunningham

2022-02-04

How do you FOIL $\left(2x-3\right)\left({x}^{2}+5x-3\right)$?

Katherine Trevino

This is where FOIL is not quite enough - Its

Pierce Carey

Alternatively, working with each of the powers of x from ${x}^{3}$ down to ${x}^{0}$ (i.e. the constant term), match the terms in the first and second bracketed expressions which will multiply to give that power of x and add them together, thus:
Given: $\left(2x-3\right)\left({x}^{2}+5x-3\right)$
${x}^{3}:2x×{x}^{2}=2{x}^{3}$
${x}^{2}:\left(2x×5x\right)+\left(-3×{x}^{2}\right)=10{x}^{2}-3{x}^{2}=7{x}^{2}$
${x}^{1}:\left(2x×-3\right)+\left(-3×5x\right)=-6x-15x=-21x$
${x}^{0}:-3×-3=9$
Added, give $2{x}^{3}+7{x}^{2}-21x+9$

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