Factorising the Quadratic

Question

The purpose of this question is to gain an explanation and understand the formula and rules factorising the quadratic ${x^2}-7x+12$:

$$ {\begin{align}{x^2}-7x+12 = {x^2}-4x-3x+12\\ = x(x-4)-3x+12\\ = x(x-4)-3(x-4)\\ = (x-4)(x-3) \end{align}} $$

looking at the line $= x(x-4)-3(x-4)\\$, the next logical step is to move the $-3$) such that:

$(x-3)(x-4)(x-4)$

Is there a proof or formula that shows why the common factor $(x-4)$ is dropped to result in?

$(x-4)(x-3)$

Answer 1

The factor isn't dropped, you apply the distribution rule

$$ac-bc=(a-b)c$$

not the "Wonderland" rule

$$ac-bc=(a-b)cc.$$

For instance,

$$4\cdot2-3\cdot2=1\cdot2\ne1\cdot2\cdot2.$$

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For an intuitive explanation, $$ac-bc=(\underbrace{c+c+c+\cdots c}_{a\text{ terms}})-(\underbrace{c+c+c+\cdots c}_{b\text{ terms}})=\underbrace{c+c+c+\cdots c}_{a-b\text{ terms}}=(a-b)c.$$

Answer 2

You cant simply 'move' the $-3$ with $x$ into a new factor; the first $x$ is being multiplied by the factor $(x-4)$, and the negative three is being multiplied by by $(x-4)$ too. $(x-3)(x-4)(x-4)$ will result in a third degree polynomial, which isn't quadratic, and so wouldn't make sense as an answer either.

You can use the helpful identity $[a(b+c)=ab+ac]$ to turn $x(x-4)-3(x-4)$ to $(x-4)(x-3)$, no dropping or 'cancelling' involved.

You could even use the quadratic formula. Use substitution and simplifying to then get an answer of $(x-4)(x-3)$.