Multiply (sic) ${(2 x - 5)}^{2}$ $(2x-5)^2$ ?

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Answer 1 · 2017-08-24T03:24:07

$4 x^{2} - 20 x + 25$ $4x^2-20x+25$

${(2 x - 5)}^{2}$ $(2x-5)^2$

[N.B. By "multiply" I assume you mean expand]

Remember: ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a-b)^2 = a^2-2ab+b^2$

We can prove this by expanding term by term as follows:

${(a - b)}^{2} = (a - b) (a - b)$ $(a-b)^2 = (a-b)(a-b)$

$= a (a - b) - b (a - b)$ $= a(a-b) - b(a-b)$

$= a^{2} - a b - b a + b^{2}$ $=a^2-ab -ba +b^2$

Since $b a = a b$ $ba = ab$

$= a^{2} - a b - a b + b^{2}$ $=a^2-ab -ab +b^2$

$= a^{2} - 2 a b + b^{2}$ $= a^2-2ab+b^2$

In our example $a = 2 x and b = 5$ $a=2x and b=5$

$:. (2x-5)^2 =(2x)^2 - 2xx 2x xx 5 +5^2$

$= 4x^2 -20x +25$

Multiply (sic) (2x-5)^2(2x−5)2(2x-5)^2 ?