I was asked to prepare a set of notes for year 12 students. These notes are going to be used by these students to prepare for the examinations. Naturally, I searched for problems to include so the students could get practice and I came across the following problem
Factorise $7x^2 + 7x - 7$ as far as possible
When making the solutions document I had difficulties in going far "as far as possible". I easily saw that $7(x^2 + x - 1)$ was one possible factorisation, but then I proceeded to see if I could factorise $x^2 + x - 1$ and it was here that I became stuck. I tried using the quadratic formula to find the roots and express $x^2 + x - 1$ as $(x_1 - root1)(x_2 - root2)$, but I couldn't seem to find such an expression. After all of that I used Wolfram Alpha and was given an irreducible factorisation of $-1/4 (-2x + \sqrt{5} -1)(2x + \sqrt{5} + 1)$. I tried to work backwards to see if could come to this factorisation, but either my lack of creativity or my terrible computation skills have left with me no solution.
If I could get a hint (maybe a solution too, but one hidden by spoiler block would be nice just so I can attempt it myself first), I'd be really thankful.
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$\begingroup$You can either use the quadratic formula or complete the square. The quadratic formula gives you the roots directly. If you don't know it, write$$ \begin {align} x^2+x-1&=\left(x^2+x+\frac 14\right)-\frac 54\\ &=\left(x+\frac 12\right)^2-\frac 54\\ &=\left(x+\frac 12+\frac {\sqrt 5}2\right)\left(x+\frac 12-\frac {\sqrt 5}2\right) \end {align} $$where you can combine the fractions if you want. From the second to the third line is seeing the difference of squares.
$\endgroup$ 4 $\begingroup$I tried using the quadratic formula to find the roots and express $x^2 + x - 1$ as $(x_1 - root1)(x_2 - root2)$, but I couldn't seem to find such an expression.
That's a good idea; perhaps something went wrong when finding those roots?
Carefully use the quadratic formule for the roots of $\color{red}{a}x^2 + \color{blue}{b}x + \color{green}{c}$, so:
$$\color{red}{1}x^2 + \color{blue}{1}x \color{green}{-1} = 0 \iff x = \frac{- \color{blue}{1} \pm \sqrt{\color{blue}{1}^2-4 \cdot \color{red}{1} \cdot (\color{green}{-1}})}{2\cdot\color{red}{1}}=\frac{-1 \pm \sqrt{5}}{2}$$So:$$x^2 + x - 1=\left(x-\left(\frac{-1 + \sqrt{5}}{2}\right)\right)\left(x-\left(\frac{-1 - \sqrt{5}}{2}\right)\right)$$You could rewrite/simplify this further to see it matches WolframAlpha's answer.
$\endgroup$ 4 $\begingroup$Write your equation as$$x^2 + x -1 = (Ax+B)(Cx+D)$$for some real numbers $A,B,C,D$. Multiplying out the right side, we find that$$x^2+x-1 = ACx^2 + (BC + AD)x - AC$$This implies that $AC = 1$, $BC + AD = 1$, and $BD = -1$. You can solve this system to return some desired values.
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