How do you factor the trinomial $x^{2} - 2 x + 1$ $x^2-2x+1$ ?

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Answer 1 · 2017-02-14T19:47:40

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This is a quadratic and, if all else fails, you revert to the Quadratic Formula .

For generalised quadratic $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ , we know, from the Quadratic Formula, that its roots are:

$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x_(1,2) = (-b pmsqrt (b^2 - 4ac))/(2a)$

In this particular case, we get:

$x_{1, 2} = \frac{2 \pm \sqrt{4 - 4}}{2} = 1$ $x_(1,2) = (2 pmsqrt (4 - 4))/(2) = 1$ , ie repeated roots

Of course we are smarter than (!) that and we will have noted at the outset that $x^{2} - 2 x + 1$ $x^2-2x+1$ can be factored as:

$x^{2} - 2 x + 1$ $x^2-2x+1$

$= (x - 1) (x - 1)$ $= (x-1)(x-1)$

Factoring, without the sometimes drudgery of the Quadratic Formula, comes down to experience.

If you see that:

$(x - α) (x - β) = x^{2} - (α + β) x + α β$ $(x-alpha)(x - beta) = x^2 - (alpha + beta)x + alpha beta$

...then you can start to look for patterns.

In this case, the patterns were:

$α β = 1$ $alpha beta = 1$
$- (α + β) = - 2$ $-(alpha + beta) = -2$ , ie $(α + β) = 2$ $(alpha + beta) = 2$ .

It follows that $α = β = 1$ $alpha = beta = 1$ .

How do you factor the trinomial x2−2x+1x^2-2x+1?