Use the Binomial Theorem to expand ${(x + 7)}^{4}$ $(x+7)^4$ and express the result in simplified form?

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Use the Binomial Theorem to expand ${(x + 7)}^{4}$ $(x+7)^4$ and express the result in simplified form?

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Answer 1 · 2018-05-17T08:48:13

$2401 + 1372 x + 294 x^{2} + 28 x^{3} + x^{4}$ $2401+1372x+294x^2+28x^3+x^4$

Explanation:

Using binomial theorem we can express ${(a + b x)}^{c}$ $(a+bx)^c$ as an expanded set of $x$ $x$ terms:
${(a + b x)}^{c} = c \sum n = 0 \frac{c!}{n! (c - n)!} a^{c - n} {(b x)}^{n}$ $(a+bx)^c=sum_(n=0)^c(c!)/(n!(c-n)!)a^(c-n)(bx)^n$

Here, we have ${(7 + x)}^{4}$ $(7+x)^4$

So, to expand we do:
$\frac{4!}{0! (4 - 0)!} 7^{4 - 0} x^{0} + \frac{4!}{1! (4 - 1)!} 7^{4 - 1} x^{1} + \frac{4!}{2! (4 - 2)!} 7^{4 - 2} x^{2} + \frac{4!}{3! (4 - 3)!} 7^{4 - 3} x^{3} + \frac{4!}{4! (4 - 4)!} 7^{4 - 4} x^{4}$ $(4!)/(0!(4-0)!)7^(4-0)x^0+(4!)/(1!(4-1)!)7^(4-1)x^1+(4!)/(2!(4-2)!)7^(4-2)x^2+(4!)/(3!(4-3)!)7^(4-3)x^3+(4!)/(4!(4-4)!)7^(4-4)x^4$

$\frac{4!}{0! (4 - 0)!} 7^{4} x^{0} + \frac{4!}{1! (4 - 1)!} 7^{3} x^{1} + \frac{4!}{2! (4 - 2)!} 7^{2} x^{2} + \frac{4!}{3! (4 - 3)!} 7 x^{3} + \frac{4!}{4! (4 - 4)!} 7^{0} x^{4}$ $(4!)/(0!(4-0)!)7^4x^0+(4!)/(1!(4-1)!)7^3x^1+(4!)/(2!(4-2)!)7^2x^2+(4!)/(3!(4-3)!)7x^3+(4!)/(4!(4-4)!)7^0x^4$

$\frac{4!}{0! 4!} 7^{4} + \frac{4!}{1! 3!} 7^{3} x + \frac{4!}{2! 2!} 7^{2} x^{2} + \frac{4!}{3! 1!} 7 x^{3} + \frac{4!}{4! 0!} x^{4}$ $(4!)/(0!4!)7^4+(4!)/(1!3!)7^3x+(4!)/(2!2!)7^2x^2+(4!)/(3!1!)7x^3+(4!)/(4!0!)x^4$

$7^{4} + 4 {(7)}^{3} x + \frac{24}{4} 7^{2} x^{2} + 4 (7) x^{3} + x^{4}$ $7^4+4(7)^3x+24/4 7^2x^2+4(7)x^3+x^4$

$2401 + 1372 x + 294 x^{2} + 28 x^{3} + x^{4}$ $2401+1372x+294x^2+28x^3+x^4$

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