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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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    $\begingroup$ To properly answer we need to know why you "couldn't seem to find such an expression", i.e. show us what you did so we can see what went awry. Otherwise we are forced to make (wild) guesses. $\endgroup$ – Bill Dubuque Feb 12 '19 at 16:09
  • $\begingroup$ @BillDubuque I didn't know if my idea was leading to my a correct solution (an irreducible factorisation). I looked at wolfram alpha's output and I doubted that I was on the right track. I may have also made some careless errors. A combination of the above I guess would an answer to why I couldn't or rather found it difficult to find a solution $\endgroup$ – Herb Feb 12 '19 at 16:14
  • $\begingroup$ @BillDubuque I didn't know if my idea was leading to my a correct solution (an irreducible factorisation). I looked at wolfram alpha's output and I doubted that I was on the right track. I may have also made some careless errors. A combination of the above I guess would an answer to why I couldn't or rather found it difficult to find a solution $\endgroup$ – Herb Feb 12 '19 at 16:14
  • $\begingroup$ It's still not clear why you couldn't compute the factorization using the quadratic formula. Precisely where did you have problems doing so? $\endgroup$ – Bill Dubuque Feb 12 '19 at 16:23
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    $\begingroup$ Do you meam this: $\,(-1/4)(-2x +a)(2x+b) = (x-a/2)(x+b/2)$? That follows by writing $\,-1/4 = (-1/2)(1/2)\,$ and distributing the $-1/2$ into $-2x+a$ and the $1/2$ into $2x+b,\,$ or, equivalently pulling a factor of $-2$ and $2$ out of each, with the goal to make the leading coef $=1$ in each linear factor. If you had shown your work we could have narrowed that down quite quickly. $\endgroup$ – Bill Dubuque Feb 12 '19 at 16:41
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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.

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  • $\begingroup$ This is quite a pretty solution. Thanks Ross. I remember similar problems where such a trick worked wonders. I guess I needed a break because I saw the solution so quickly afterwards $\endgroup$ – Herb Feb 12 '19 at 16:41
  • $\begingroup$ The quadratic formula is derived in exactly this manner. It is essentially a generic completion of the square. Of course sometimes the roots are complex. $\endgroup$ – Ross Millikan Feb 12 '19 at 16:44
  • $\begingroup$ I do remember that. This reminds me that you can learn a good deal from proofs. $\endgroup$ – Herb Feb 12 '19 at 16:50
  • $\begingroup$ @BillDubuque: No worries. $\endgroup$ – Ross Millikan Feb 12 '19 at 17:08
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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.

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    $\begingroup$ Who is down-voting this (and the other) answer(s); and, more importantly, what for...? $\endgroup$ – StackTD Feb 12 '19 at 16:13
  • $\begingroup$ Thanks for the quick reply $\endgroup$ – Herb Feb 12 '19 at 16:38
  • $\begingroup$ I managed to get it after a small break $\endgroup$ – Herb Feb 12 '19 at 16:38
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    $\begingroup$ You're welcome and well done. $\endgroup$ – StackTD Feb 12 '19 at 16:39
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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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