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# Secondary Magazine - Issue 125: Sixth Sense

Created on 10 September 2015 by ncetm_administrator
Updated on 01 October 2015 by ncetm_administrator

# Sixth Sense

One of the key skills you’ll soon be teaching your new year 12 Pure Maths group is how to factorise cubics (and higher order polynomials).

If you’ve already shown your students the “field method” (see Building Bridges, Issue 119), so much the better – if not, now is the time to convince them of its value. They might well have seen in KS2 how to multiply 13 by 82:

 × $\inline \dpi{80} \fn_jvn \small 10$ $\inline \dpi{80} \fn_jvn \small 3$ $\inline \dpi{80} \fn_jvn \small 80$ $\inline \dpi{80} \fn_jvn \small 800$ $\inline \dpi{80} \fn_jvn \small 240$ $\inline \dpi{80} \fn_jvn \small 2$ $\inline \dpi{80} \fn_jvn \small 20$ $\inline \dpi{80} \fn_jvn \small 6$ $\inline \dpi{80} \fn_jvn \small 1066$

and then in KS3 developed the method to expand $\inline \dpi{80} \fn_jvn \small \left ( 2x+3 \right )\left ( x-5 \right )$:

 × $\inline \dpi{80} \fn_jvn \small 2x$ $\inline \dpi{80} \fn_jvn \small 3$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small 2x^{2}$ $\inline \dpi{80} \fn_jvn \small 3x$ $\inline \dpi{80} \fn_jvn \small -5$ $\inline \dpi{80} \fn_jvn \small -10x$ $\inline \dpi{80} \fn_jvn \small -15$ $\inline \dpi{80} \fn_jvn \small 2x^{2}-7x-15$

and then in KS4 used the same model to factorise $\dpi{80} \fn_jvn \small \dpi{80} \fn_jvn \small x^{2}-7x-30$:

FIRST ATTEMPT (picking two integers that multiply to equal -30)

 × $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small -5$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{2}$ $\inline \dpi{80} \fn_jvn \small -5x$ $\inline \dpi{80} \fn_jvn \small 6$ $\inline \dpi{80} \fn_jvn \small 6x$ $\inline \dpi{80} \fn_jvn \small -30$ $\inline \dpi{80} \fn_jvn \small x^{2}+x-30$

SECOND ATTEMPT (because that didn’t quite work)

 × $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small 10$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{2}$ $\inline \dpi{80} \fn_jvn \small 10x$ $\inline \dpi{80} \fn_jvn \small -3$ $\inline \dpi{80} \fn_jvn \small -3x$ $\inline \dpi{80} \fn_jvn \small -30$ $\inline \dpi{80} \fn_jvn \small x^{2}+7x-30$

GOT IT: $\inline \dpi{80} \fn_jvn \small x^{2}+7x-30\equiv \left ( x+10 \right )\left ( x-3 \right )$.

Before attempting to factorise any cubics, it is worth using a field to expand the product of three factors. e.g. Expand and simplify $\inline \dpi{80} \fn_jvn \small \left ( 2x+3 \right )\left ( x-5 \right )\left ( x+2 \right )$.

Step 1 – expand the first two factors (see above)

Step 2 – multiply the answer by the third factors:

 × $\inline \dpi{80} \fn_jvn \small 2x^{2}$ $\inline \dpi{80} \fn_jvn \small -7x$ $\inline \dpi{80} \fn_jvn \small -15$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small 2x^{3}$ $\inline \dpi{80} \fn_jvn \small -7x^{2}$ $\inline \dpi{80} \fn_jvn \small -15x$ $\inline \dpi{80} \fn_jvn \small 2$ $\inline \dpi{80} \fn_jvn \small 4x^{2}$ $\inline \dpi{80} \fn_jvn \small -14x$ $\inline \dpi{80} \fn_jvn \small -30$ $\inline \dpi{80} \fn_jvn \small 2x^{3}-3x^{2}-29x-30$

Practice will help students become fluent with this procedure. Now let’s look at how to use this to factorise a cubic when we are given one of the factors.
e.g. Factorise $\inline \dpi{80} \fn_jvn \small x^{3}+x^{2}-16x+20$ given that $\inline \dpi{80} \fn_jvn \small x-2$ is a factor.

Step 1 – present the known information

 × $\inline \dpi{80} \fn_jvn \small ?x^{2}$ $\inline \dpi{80} \fn_jvn \small ?x$ $\inline \dpi{80} \fn_jvn \small ?$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small -2$ $\inline \dpi{80} \fn_jvn \small x^{3}+x^{2}-16x+20$

Step 2 – to have an $\inline \dpi{80} \fn_jvn \small x^{3}$ term, the first term of the quadratic must be just $\inline \dpi{80} \fn_jvn \small x^{2}$, so:

 × $\inline \dpi{80} \fn_jvn \small x^{2}$ $\inline \dpi{80} \fn_jvn \small ?x$ $\inline \dpi{80} \fn_jvn \small ?$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{3}$ $\inline \dpi{80} \fn_jvn \small -2$ $\inline \dpi{80} \fn_jvn \small -2x^{2}$ $\inline \dpi{80} \fn_jvn \small x^{3}+x^{2}-16x+20$

Step 3 – consequently, we have a new term, $\dpi{80} \fn_jvn \small -2x^{2}$, but eventually we want just $\dpi{80} \fn_jvn \small x^{2}$, so we need to create another term, $\dpi{80} \fn_jvn \small 3x^{2}$:

 × $\inline \dpi{80} \fn_jvn \small x^{2}$ $\inline \dpi{80} \fn_jvn \small ?x$ $\inline \dpi{80} \fn_jvn \small ?$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{3}$ $\inline \dpi{80} \fn_jvn \small 3x^{2}$ $\inline \dpi{80} \fn_jvn \small -2$ $\inline \dpi{80} \fn_jvn \small -2x^{2}$ $\inline \dpi{80} \fn_jvn \small x^{3}+x^{2}-16x+20$

Step 4 – so the coefficient of the $\dpi{80} \fn_jvn \small x$ term in the quadratic factor must be $\dpi{80} \fn_jvn \small 3$, and we can fill in another patch in the field:

 × $\inline \dpi{80} \fn_jvn \small x^{2}$ $\inline \dpi{80} \fn_jvn \small 3x$ $\inline \dpi{80} \fn_jvn \small ?$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{3}$ $\inline \dpi{80} \fn_jvn \small 3x^{2$ $\inline \dpi{80} \fn_jvn \small -2$ $\inline \dpi{80} \fn_jvn \small -2x^{2}$ $\inline \dpi{80} \fn_jvn \small -6x$ $\inline \dpi{80} \fn_jvn \small x^{3}+x^{2}-16x+20$

Step 5 (like step 3) – we now have a new term, $\dpi{80} \fn_jvn \small -6x$, but ultimately we want the term to be $\dpi{80} \fn_jvn \small -16x$, so we have no choice on what to fill in next:

 × $\inline \dpi{80} \fn_jvn \small x^{2}$ $\inline \dpi{80} \fn_jvn \small 3x$ $\inline \dpi{80} \fn_jvn \small ?$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{3}$ $\inline \dpi{80} \fn_jvn \small 3x^{2$ $\inline \dpi{80} \fn_jvn \small -10x$ $\inline \dpi{80} \fn_jvn \small -2$ $\inline \dpi{80} \fn_jvn \small -2x^{2}$ $\inline \dpi{80} \fn_jvn \small -6x$ $\inline \dpi{80} \fn_jvn \small x^{3}+x^{2}-16x+20$

Step 6 – thus the final “$\dpi{80} \fn_jvn \small ?$” needs to be “$\dpi{80} \fn_jvn \small -10$” to create the desired term:

 × $\inline \dpi{80} \fn_jvn \small x^{2}$ $\inline \dpi{80} \fn_jvn \small 3x$ $\inline \dpi{80} \fn_jvn \small -10$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{3}$ $\inline \dpi{80} \fn_jvn \small 3x^{2$ $\inline \dpi{80} \fn_jvn \small -10x$ $\inline \dpi{80} \fn_jvn \small -2$ $\inline \dpi{80} \fn_jvn \small -2x^{2}$ $\inline \dpi{80} \fn_jvn \small -6x$ $\inline \dpi{80} \fn_jvn \small 20$ $\inline \dpi{80} \fn_jvn \small x^{3}+x^{2}-16x+20$

This is very satisfying, because -10 multiplied by -2 is +20, so we must have factorised the cubic correctly. This is why I would discourage students from trying to “hit” the constant term first: if they get to the end and the numerical multiplication works then it is highly likely they’ve completed the process correctly; if it doesn’t, then they immediately know that they haven’t! So we now know that:

$\inline \dpi{80} \fn_jvn \small x^{3}+x^{2}-16x+20\equiv \left ( x-2 \right )\left ( x^{2}+3x-10 \right )$

In this example, we can factorise the quadratic factor, so:

$\inline \dpi{80} \fn_jvn \small x^{3}+x^{2}-16x+20\equiv \left ( x-2 \right )^{2}\left ( x+5 \right )$

This process looks long-winded in print or on a screen, but clearly on a whiteboard (classroom or handheld) the field only needs drawing once and judicious use of coloured pens can make the whole example compact and clear!

All that remains is to factorise a cubic when we are not given a factor initially. In my experience, students are easily convinced (informally) of the factor theorem given their prior experience of solving quadratic equations: they will, with perhaps a little prompting, recall that “if one of the factors is $\inline \dpi{80} \fn_jvn \small \left ( x+7 \right )$ then one of the roots is $\inline \dpi{80} \fn_jvn \small x= -7$, and this means that substituting into the original quadratic expression gives an output value of 0”.

Therefore, reversing the argument, when given a quadratic / cubic / higher order polynomial expression, if we can find a “magic number” that, when substituted, makes the expression equal 0, we have found a root and hence a factor. For example, to factorise $\inline \dpi{80} \fn_jvn \small 2x^{3}-7x^{2}+11x-10$ the students first need to determine that the cubic expression is equal to 0 when $\inline \dpi{80} \fn_jvn \small x= 2$, and so one of the factors is $\inline \dpi{80} \fn_jvn \small \left ( x-2 \right )$. They now proceed with the field method. This is preferable to seeking a second “magic number” in general because (a) we may have a repeated factor as was the case above, and no simple (without calculus) way of seeing this; or (b) the other factor may be an irreducible polynomial, as is the case here.

Students’ conceptual understanding of the method can be deepened by asking them “what happens if we try and factorise $\inline \dpi{80} \fn_jvn \small x^{3}-4x^{2}+2x+7$ using $\inline \dpi{80} \fn_jvn \small \left ( x-3 \right )$ as our factor?” Let’s pick this up next month.

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02 October 2015 07:54
I love it to, much to the chagrin of one of my colleagues, who thinks that the Chinese method should be the go to method for lower secondary. My thing with it, is not just with the move to algebra, but also it can form a strong link with columnar long multiplication, as prescribed by the new primary curriculum.