Notes: `(4x - 1)^3 - 2(4x - 1)^4` Use the Binomial Theorem to expand and simplify the expression.

Wednesday, January 25, 2012

$\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}-{2}{{\left({4}{x}-{1}\right)}}^{{4}}$ Use the Binomial Theorem to expand and simplify the expression.

You need first to factorize $\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}$ , such that:

$\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}-{2}{{\left({4}{x}-{1}\right)}}^{{4}}={{\left({4}{x}-{1}\right)}}^{{3}}{\left({1}-{2}{\left({4}{x}-{1}\right)}\right)}$

$\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}-{2}{{\left({4}{x}-{1}\right)}}^{{4}}={{\left({4}{x}-{1}\right)}}^{{3}}{\left({1}-{8}{x}+{2}\right)}$

$\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}-{2}{{\left({4}{x}-{1}\right)}}^{{4}}={{\left({4}{x}-{1}\right)}}^{{3}}{\left({3}-{8}{x}\right)}$

You need to use the binomial formula, such that:

$\displaystyle{{\left({x}+{y}\right)}}^{{n}}={\sum_{{{k}={0}}}^{{n}}}{\left(\matrix{{n}\\{k}}\right)}{{x}}^{{{n}-{k}}}{{y}}^{{k}}$

You need to replace 4x for x, 1 for y and 3 for n, such that:

$\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}={3}{C}{0}{{\left({4}{x}\right)}}^{{3}}+{3}{C}{1}{{\left({4}{x}\right)}}^{{2}}\cdot{{\left(-{1}\right)}}^{{1}}+{3}{C}{2}{\left({4}{x}\right)}\cdot{{\left(-{1}\right)}}^{{2}}+{3}{C}{3}{{\left(-{1}\right)}}^{{3}}$

By definition, nC0 = nCn = 1, hence $\displaystyle{3}{C}{0}={3}{C}{3}={1}.$

By definition nC1 = nC(n-1) = n, hence $\displaystyle{3}{C}{1}={3}{C}{2}={3}.$

$\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}={64}{{x}}^{{3}}-{48}{{x}}^{{2}}+{12}{x}-{1}$

Replacing the binomial expansion $\displaystyle{64}{{x}}^{{3}}-{48}{{x}}^{{2}}+{12}{x}-{1}$ for $\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}$ yields:

$\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}{\left({3}-{8}{x}\right)}={\left({64}{{x}}^{{3}}-{48}{{x}}^{{2}}+{12}{x}-{1}\right)}{\left({3}-{8}{x}\right)}$

$\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}{\left({3}-{8}{x}\right)}={\left({192}{{x}}^{{3}}-{512}{{x}}^{{4}}-{144}{{x}}^{{2}}+{384}{{x}}^{{3}}+{36}{x}-{96}{{x}}^{{2}}-{3}+{8}{x}\right)}$

Grouping the like terms yields:

$\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}{\left({3}-{8}{x}\right)}={\left(-{512}{{x}}^{{4}}+{576}{{x}}^{{3}}-{240}{{x}}^{{2}}+{44}{x}-{3}\right)}$

Hence, expanding and simplifying the expression yields $\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}-{2}{{\left({4}{x}-{1}\right)}}^{{4}}={\left(-{512}{{x}}^{{4}}+{576}{{x}}^{{3}}-{240}{{x}}^{{2}}+{44}{x}-{3}\right)}.$

Notes

Wednesday, January 25, 2012

$\displaystyle{{\left({4}{x}-{1}\right)}}^{{3}}-{2}{{\left({4}{x}-{1}\right)}}^{{4}}$ Use the Binomial Theorem to expand and simplify the expression.

No comments:

Post a Comment

How does the choice of details set the tone of the sermon?