How do you simplify ${(x - 3)}^{2} - 4 (x - 3) + 3$ $(x-3)^2-4(x-3)+3$ ?

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Answer 1 · 2018-04-13T10:57:09

$(x - 6) (x - 4)$ $(x-6)(x-4)$

${(x - 3)}^{2} - 4 (x - 3) + 3$ $(x-3)^2-4(x-3)+3$

$((x - 3) - 3) ((x - 3) - 1)$ $((x-3)-3)((x-3)-1)$

$(x - 6) (x - 4)$ $(x-6)(x-4)$

Answer 2 · 2018-04-16T07:47:49

Simplified gives $x^{2} - 10 x + 24$ $x^2 -10x+24$

and factorised give: $(x - 6) (x - 4)$ $(x-6)(x-4)$

To simplify we need to multiply out the brackets:

${(x - 3)}^{2} - 4 (x - 3) + 3$ $(x-3)^2 -4(x-3) +3$

$= x^{2} - 6 x + 9 - 4 x + 12 + 3$ $=x^2-6x+9-4x+12+3$

$= x^{2} - 10 x + 24$ $=x^2 -10x+24$

However, I suspect that the question was meant to be to factorise..

We could multiply out as done above and then factorise from this, but let's regard $(x - 3)$ $(x-3)$ as a variable, $p$ $p$ .

${(x - 3)}^{2} - 4 (x - 3) + 3$ $(x-3)^2 -4(x-3) +3$

$\to p^{2} - 4 p + 3$ $rarr p^2 -4p +3$

$= (p - 3) (p - 1)$ $=(p-3)(p-1)$

$\to (x - 3 - 3) (x - 3 - 1)$ $rarr(x-3-3)(x-3-1)$

$= (x - 6) (x - 4)$ $=(x-6)(x-4)$

How do you simplify (x-3)^2-4(x-3)+3(x−3)2−4(x−3)+3(x-3)^2-4(x-3)+3?