View Full Version : Power Of Sums: expand (x + 5)^3

Richay

06-11-2006, 06:23 PM

(x + 5)^3

I looked it up on google and every site had weird formulas that made no sense to me. How do I solve this?

\L

\begin{array}{l}

\left( {a + b} \right)^3 = a^3 + 3a^2 b + 3ab^2 + b^3 \\

\mbox{in general}\

(a + b)^n = \sum\limits_{k = 0}^n {\left( \begin{array}{c}

n \\

k \\

\end{array} \right)a^{n - k} b^k } \\

\end{array}

Mrspi

06-12-2006, 09:20 AM

(x + 5)^3

I looked it up on google and every site had weird formulas that made no sense to me. How do I solve this?

You could do this the "elementary" way:

(x + 5)<sup>3</sup> means (x + 5)(x + 5)(x + 5)

Start by multiplying two of the factors together, then take that product times the third factor....

Richay

06-12-2006, 09:17 PM

Hmmm

So would 5 be the factors? What about x?

stapel

06-12-2006, 09:30 PM

So would 5 be the factors? What about x?

No. The factors are the ones that were provided to you:

. . . . .(x + 5)<sup>3</sup> = (x + 5)(x + 5)(x + 5)

The factors (the things multiplied) are x + 5. Now multiply the factors together, and simplify to get the required polynomial.

Eliz.

\L

\begin{array}{rcl}

\left( {x + 5} \right)^3 & = & x^3 + 3x^2 \left( 5 \right) + 3x\left( 5 \right)^2 + \left( 5 \right)^3 \\

& = & x^3 + 15x^2 + 75x + 125 \\

\end{array}

leilsilver

06-12-2006, 11:47 PM

Pascall's Triangle is a really helpful tool in factoring things like (x+y)^n. Like how pka showed you on the first line where:

(x+5)^3=x^3+3x^2(5)+3x(5)+5^3

The constants 1 3 3 1 comes from the fourth line in the triangle: where you would raise something to the thirdth power.

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