How do you multiply ${(2 r + 9 s)}^{2}$ $(2r+9s)^2$ ?

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How do you multiply ${(2 r + 9 s)}^{2}$ $(2r+9s)^2$ ?

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Answer 1 · 2016-08-29T13:51:08

$4 r^{2} + 36 r s + 81 s^{2}$ $4r^2+36rs+81s^2$

Explanation:

Consider ${(2 r + 9 s)}^{2} = (2 r + 9 s) (2 r + 9 s)$ $(2r+9s)^2=(2r+9s)(2r+9s)$

We must ensure that each term in the 2nd bracket is multiplied by each term in the first bracket.This can be achieved as follows.

$(2 r + 9 s) (2 r + 9 s)$ $(color(red)(2r+9s))(2r+9s)$

$= 2 r (2 r + 9 s) + 9 s (2 r + 9 s)$ $=color(red)(2r)(2r+9s)+color(red)(9s)(2r+9s)$

distribute the brackets : $4 r^{2} + 18 r s + 18 r s + 81 s^{2}$ $4r^2+18rs+18rs+81s^2$

and simplifying to obtain : $4 r^{2} + 36 r s + 81 s^{2}$ $4r^2+36rs+81s^2$

$\Rightarrow {(2 r + 9 s)}^{2} = 4 r^{2} + 36 r s + 81 s^{2}$ $rArr(2r+9s)^2=4r^2+36rs+81s^2$
$----------------------------------------------------------$ $"----------------------------------------------------------"$
Alternatively there is the FOIL method.

$F \to First terms$ $Fto" First terms"$
$O \to Outer terms$ $Oto" Outer terms"$
$I \to Inner terms$ $Ito" Inner terms"$
$L \to Last terms$ $Lto" Last terms"$

F-multiply the first terms in each bracket together.
O-multiply the outer terms together.
I-multiply the inner terms together.
L-multiply the last terms together.

$\Rightarrow (2 r + 9 s) (2 r + 9 s)$ $rArr(2r+9s)(2r+9s)$

$= (2 r \times 2 r) + (2 r \times 9 s) + (2 r \times 9 s) + (9 s \times 9 s)$ $=(2rxx2r)+(2rxx9s)+(2rxx9s)+(9sxx9s)$

$= 4 r^{2} + 18 r s + 18 r s + 81 s^{2} = 4 r^{2} + 36 r s + 81 s^{2}$ $=4r^2+18rs+18rs+81s^2=4r^2+36rs+81s^2$

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How do you multiply ${(2 r + 9 s)}^{2}$ $(2r+9s)^2$ ?

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