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# Secondary Magazine - Issue 126: Sixth Sense

Created on 12 October 2015 by ncetm_administrator
Updated on 22 October 2015 by ncetm_administrator

# Sixth Sense

Last month we looked at factorising cubics (and, by extension, higher order polynomials) by adapting the multiplication model of “farmers’ fields”. I ended by wondering what happens if we try to “factorise” $\inline \dpi{80} \fn_jvn \small x^{3}-4x^{2}+2x+7$ using  as our “factor”.

As before, we can express the given information as a field  and see what happens:

 $\inline \dpi{80} \fn_jvn \small x$ $\dpi{80} \fn_jvn \small ?x^{2}$ $\dpi{80} \fn_jvn \small ?x$ $\dpi{80} \fn_jvn \small ?$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{3}$ $\inline \dpi{80} \fn_jvn \small -3$ $\inline \dpi{80} \fn_jvn \small x^{3}-4x^{2}+2x+7$

The colour-coded diagram below shows the following steps:

Step 1 – to get the $\dpi{80} \fn_jvn \small {\color{Red} x^{3}}$ term, we just need $\dpi{80} \fn_jvn \small {\color{Red} 1x^{2}}$; this generates the $\dpi{80} \fn_jvn \small {\color{Red} -3x^{2}}$ term.

Step 2 – since we want a total quadratic term of $\dpi{80} \fn_jvn \small {\color{DarkGreen} -4x^{2}}$, we have to have a $\dpi{80} \fn_jvn \small {\color{DarkGreen} -x}$ term in our quadratic “factor”; this also generates the $\dpi{80} \fn_jvn \small {\color{DarkGreen} 3x}$ term.

Step 3 – since we want a total linear term of $\dpi{80} \fn_jvn \small {\color{Blue} +2x}$, the final entry in the top row must be $\dpi{80} \fn_jvn \small {\color{Blue} -1}$

 $\inline \dpi{80} \fn_jvn \small x$ $\dpi{80} \fn_jvn \small \dpi{80} \fn_jvn \small {\color{Red} 1}x^{2}$ $\dpi{80} \fn_jvn \small {\color{DarkGreen}-1}x$ $\dpi{80} \fn_jvn \small {\color{Blue} -1}$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{3}$ $\dpi{80} \fn_jvn \small {\color{DarkGreen} -x^{2}}$ $\dpi{80} \fn_jvn \small {\color{Blue} -x}$ $\inline \dpi{80} \fn_jvn \small -3$ $\dpi{80} \fn_jvn \small {\color{Red} -3x^{2}}$ $\dpi{80} \fn_jvn \small {\color{DarkGreen} 3x}$ $\dpi{80} \fn_jvn \small {\color{Blue} 3}$ $\inline \dpi{80} \fn_jvn \small x^{3}-4x^{2}+2x+7$

Thus our attempt to factorise hasn’t worked – we’ve ended up with a constant term of “$\dpi{80} \fn_jvn \small 3$” when a correct factorisation would have given us “$\dpi{80} \fn_jvn \small \dpi{80} \fn_jvn \small 7$”. However, we can instead deduce that:

$\dpi{80} \fn_jvn \small x^{3}-4x^{2}+2x+7\equiv \left ( x-3 \right )\left ( x^{2}-x-1 \right )+4$

or equivalently,

$\dpi{80} \fn_jvn \small \frac{x^2-4x^2+2x+7{}{}}{x-3}\equiv x^{2}-x-1+\frac{4}{x-3}$

To enable students to use the right language, I often use a numerical analogy, such as “what’s $\dpi{80} \fn_jvn \small 30$ divided by $\dpi{80} \fn_jvn \small 7$?” Since

$\dpi{80} \fn_jvn \small 30=7\times 4+2$

the answer is sometimes given as “$\dpi{80} \fn_jvn \small 4$ remainder $\dpi{80} \fn_jvn \small 2$”, or:

$\dpi{80} \fn_jvn \small \frac{30}{7}=4+\frac{2}{7}=4\frac{2}{7}$

Your students will have known since primary school about remainders; they may not know that the “$\dpi{80} \fn_jvn \small 4$” in this calculation is called “the quotient”. Extending this language to our algebraic example, they can now say that “$\dpi{80} \fn_jvn \small x^{3}-4x^{2}+2x+7$  divided by $\dpi{80} \fn_jvn \small x-3$ gives a quotient of $\dpi{80} \fn_jvn \small x^{2}-x-1$ and a remainder of $\dpi{80} \fn_jvn \small 4$”.

There’s an important parallel here between numerical top-heavy fractions (if the numerator is greater than or equal to the denominator we can simplify the division, as we did with $\dpi{80} \fn_jvn \small 30\div 7$) and algebraic ones (if the degree of the numerator is greater than or equal to the degree of the denominator we can simplify the division to find the quotient and remainder).

This seems like a good time for some practice: I’d ask students to find quotients and remainders using more fields. Just think how their algebraic fluency will improve as they do so! It’s well worth their considering a non-linear divisor: For example, Find the quotient and remainder when $\dpi{80} \fn_jvn \small x^{4}+7x^{3}-2x+8$ is divided by $\dpi{80} \fn_jvn \small x^{2}+3x+5$, leads to this field (where, as before, each column from left to right gets filled in, in turn, from top to bottom):

 $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{2}$ $\dpi{80} \fn_jvn \small 4x$ $\inline \dpi{80} \fn_jvn \small -17$ $\inline \dpi{80} \fn_jvn \small x$ $\inline \dpi{80} \fn_jvn \small x^{4}$ $\inline \dpi{80} \fn_jvn \small 4x^{3}$ $\inline \dpi{80} \fn_jvn \small -17x^{2}$ $\inline \dpi{80} \fn_jvn \small 3x$ $\inline \dpi{80} \fn_jvn \small 3x^{3}$ $\inline \dpi{80} \fn_jvn \small 12x^{2}$ $\inline \dpi{80} \fn_jvn \small -51x$ $\inline \dpi{80} \fn_jvn \small 5$ $\inline \dpi{80} \fn_jvn \small 5x^{2}$ $\inline \dpi{80} \fn_jvn \small 20x$ $\inline \dpi{80} \fn_jvn \small -85$ $\inline \dpi{80} \fn_jvn \small x^{4}+7x^{3}-2x+8$

From which they conclude that

$\dpi{80} \fn_jvn \small x^{4}+7x^{3}-2x+8\equiv \left ( x^{2}+3x+5 \right )\left ( x^{2}+4x-17 \right )+29x+93$

and so they say that the quotient is $\dpi{80} \fn_jvn \small x^{2}+4x-17$ and that the remainder is $\dpi{80} \fn_jvn \small 29x+93$.

Establishing the Remainder Theorem follows naturally. Recall that:

$\dpi{80} \fn_jvn \small x^{3}-4x^{2}+2x+7\equiv \left ( x-3 \right )\left ( x^{2}-x-1 \right )+4$

Since this is an identity, it is true for all values of $\dpi{80} \fn_jvn \small x$. In particular, the students can notice that substituting $\dpi{80} \fn_jvn \small x=3$ into the original expression (the dividend, on the left-hand side) will have to give a value of $\dpi{80} \fn_jvn \small 4$, because the right-hand side will become $\dpi{80} \fn_jvn \small 0\times ?+4$ – and they should check that it does.

Having by now seen lots of examples, your students should be confident in making the claim that if they attempt to divide some polynomial $\dpi{80} \fn_jvn \small f\left ( x \right )$ by a linear expression $\dpi{80} \fn_jvn \small \left ( x-a \right )$ they will end up with $\dpi{80} \fn_jvn \small f\left ( x \right )=\left ( x-a \right )q\left ( x \right )+r$, where (by experience) the degree of $\dpi{80} \fn_jvn \small q$ is one less than the degree of $\dpi{80} \fn_jvn \small f$, and $\dpi{80} \fn_jvn \small r$ is constant. Thus they can substitute $\dpi{80} \fn_jvn \small x=a$ into both sides to see that $\dpi{80} \fn_jvn \small f\left ( a \right )=r$ by necessity. Do make sure that they learn to quote this result: the importance of “division by a field” is that it gives students conceptual understanding of division, but it’s not at all a procedurally fluent or efficient way to find the value of the constant remainder. The Factor Theorem, that $\dpi{80} \fn_jvn \small f\left ( a \right )=0\Leftrightarrow \left ( x-a \right )$ is a factor of $\dpi{80} \fn_jvn \small f\left ( x \right )$, is just a special case of the Remainder Theorem; again, students need to learn to quote this, not deduce it “from scratch” every time.

Division by quadratic divisors is usually a Further Maths topic, but it provides a good opportunity for single Maths students to deepen their confidence of the reasoning they’ve developed, and thus a question such as

When the polynomial $\dpi{80} \fn_jvn \small f\left ( x \right )$ is divided by $\dpi{80} \fn_jvn \small x-3$ the remainder is $\dpi{80} \fn_jvn \small 8$. When $\dpi{80} \fn_jvn \small f\left ( x \right )$ is divided by $\dpi{80} \fn_jvn \small x+1$ the remainder is $\dpi{80} \fn_jvn \small 12$. When $\dpi{80} \fn_jvn \small f\left ( x \right )$ is divided by $\dpi{80} \fn_jvn \small \left ( x-3 \right )\left ( x+1 \right )$ the quotient is $\dpi{80} \fn_jvn \small q\left ( x \right )$ and the remainder is $\dpi{80} \fn_jvn \small rx+s$. Find $\dpi{80} \fn_jvn \small r$ and $\dpi{80} \fn_jvn \small s$.

is well worth posing to them. Given all the examples they’ve considered, I’d expect that my students would be able to generalise from the linear case and produce a solution along the lines of

$\dpi{80} \fn_jvn \small f\left ( x \right )=\left ( x-3 \right )\left ( x+1 \right )q\left ( x \right )+rx + s$

$\dpi{80} \fn_jvn \small f\left ( 3 \right )=8\Rightarrow 3r+s=8$

$\dpi{80} \fn_jvn \small f\left ( -1 \right )=12\Rightarrow -r+s=12$

Hence $\dpi{80} \fn_jvn \small r=-1, s=11$.

If you’re a frequent user of Geogebra, I’m sure you’ll have already envisaged the potential for presenting the “field routine”. Bernard Murphy (MEI Programme Leader) got in touch after last month’s article to share his polynomial-division file, which allows you to choose your coefficients and your divisor -

- and then runs through the process as you move the slider at the bottom of the screen:

Very neat!

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