How do you factor $16 x^{2} - 40 x + 25$ $16x^2 -40x +25$ ?

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Answer 1 · 2016-05-22T16:07:20

$(4 x - 5) (4 x - 5) = {(4 x - 5)}^{2}$ $(4x-5)( 4x-5)=(4x-5)^2$

This is a pretty straight forward question.

We know that ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $color(red)((a-b)^2=a^2-2ab+b^2)$

Compare $a^{2} - 2 a b + b^{2}$ $a^2-2ab+b^2$ to $16 x^{2} - 40 x + 25$ $16x^2-40x+25$

Lets look at $a^{2}$ $a^2$ and $b^{2}$ $b^2$

$a^{2} = 16 x^{2}$ $a^2=16x^2$ and $b^{2} = 25$ $b^2=25$

$\Rightarrow a = 4 x$ $=>a=4x$ and $b = 5$ $b=5$

Then, $2 a b = 2 \times 4 x \times 5 = 40 x$ $2ab=2xx4x xx5 = 40x$ , as given in the expression.

Therefore,

$16 x^{2} - 40 x + 25$ $16x^2-40x+25$

$= {(4 x)}^{2} - (2 \times 4 x \times 5) + 5^{2}$ $=(4x)^2 - (2 xx 4x xx5) + 5^2$

$= {(4 x - 5)}^{2}$ $=(4x-5)^2$

The factors are $4 x - 5, 4 x - 5$ $4x-5, 4x-5$

How do you factor 16x^2 -40x +2516x2−40x+2516x^2 -40x +25?