How do you factor $z^{2} - 20 z + 100$ $z^2-20z+100$ ?

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Answer 1 · 2018-03-29T12:36:49

See process below

There are several ways to solve

1.- Using cuadratic formula

$a z^{2} + b z + c = 0$ $az^2+bz+c=0$ and his solutions $z = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $z=(-b+-sqrt(b^2-4ac))/(2a)$

In our case

$a = 1; b = - 20 and c = 100$ $a=1; b=-20 and c=100$

$z = \frac{20 \pm \sqrt{400 - 400}}{2} = \frac{20}{2} = 10$ $z=(20+-sqrt(400-400))/2=20/2=10$ that's double root, so

$(z - 10) (z - 10) = {(z - 10)}^{2}$ $(z-10)(z-10)=(z-10)^2$

2.- Using notable identities. In this case

${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a-b)^2=a^2-2ab+b^2$ we have $a = z$ $a=z$ and $b = 10$ $b=10$

${(z - 10)}^{2} = (z - 10) (z - 10) = z^{2} - 2 \cdot 10 \cdot z + 10^{2} = z^{2} - 20 z + 100$ $(z-10)^2=(z-10)(z-10)=z^2-2·10·z+10^2=z^2-20z+100$

Answer 2 · 2018-03-29T12:38:18

$(z - 10) (z - 10)$ $(z-10)(z-10)$

By sum and product Means that two numbers whose sum is - $20$ $20$ and product is 100

= $z^{2} - 10 z - 10 z + 100$ $z^2-10z-10z+100$

= $z (z - 10) - 10 (z - 10)$ $z(z-10)-10(z-10)$

= $(z - 10) (z - 10)$ $(z-10)(z-10)$

Hope this helps!

How do you factor z2−20z+100z^2-20z+100?