(a)  A factor of \(2x+3\) corresponds to a root of \(-\frac{3}{2},\) so we can use synthetic division like this:

\(-\small\frac{3}{2}\normalsize\)	\(2\)	\(3\)	\(-2\)	\(-3\)
	\(\downarrow\)	\(-3\)	\(0\)	\(3\)
	\(2\)	\(0\)	\(-2\)	\(0\)

The remainder is \(0\) so \(2x+3\) is a factor.

(b)  It is important to understand that the synthetic division in part (a) divided the cubic by \((x+\frac{3}{2}),\) not \((2x+3).\)

Now we consider the quotient \(2x^2-2.\) It factorises as \(2(x^2-1)\) which further factorises to \(2(x-1)(x+1).\)

So the full factorisation is \((x+\frac{3}{2})(2)(x-1)(x+1),\) which is much neater if we multiply the \(2\) by the first factor and present our final answer as \((2x+3)(x-1)(x+1).\)

A cubic polynomial will factorise into either the product of three linear factors or one linear factor times an irreducible quadratic factor. In this question, we haven't been prompted by a part (a) asking us to verify one of the factors. So we have to start from scratch.

Thankfully, there are a few easy tricks that should let us find the first factor and use it to factorise the entire polynomial.

We always test \(x=1\) first. Substituting \(x=1\) into the cubic, we obtain \(2(1^3)+9(1^2)-6(1)-5\) \(=2+9-6-5=0.\) It would have been quicker here just to add the coefficients. Anyway, we're in luck because it comes out to be \(0.\) So by the factor theorem, \(x=1\) is a root and so \(x-1\) is a factor.

Now we can use synthetic division to find the quotient after division by \(x-1.\)

\(1\)	\(2\)	\(9\)	\(-6\)	\(-5\)
	\(\downarrow\)	\(2\)	\(11\)	\(5\)
	\(2\)	\(11\)	\(5\)	\(0\)

The quotient of \(2x^2+11x+5\) factorises to \((x+5)(2x+1)\) using Nat 5 methods.

So the full factorisation is \((x-1)(x+5)(2x+1).\)

(a)  A factor of \(x+2\) corresponds to a root of \(-2,\) so we can use synthetic division like this:

\(-2\)	\(2\)	\(9\)	\(-1\)	\(-18\)	\(8\)
	\(\downarrow\)	\(-4\)	\(-10\)	\(22\)	\(-8\)
	\(2\)	\(5\)	\(-11\)	\(4\)	\(0\)

The remainder is \(0\) so \(x+2\) is a factor.

(b)  Now we consider the quotient \(2x^3+5x^2-11x+4.\) The coefficients \(2+5-11+4=0\) so we're in luck: \(x=1\) is another root, corresponding to the factor \(x-1.\) So we need to divide again:

\(1\)	\(2\)	\(5\)	\(-11\)	\(4\)
	\(\downarrow\)	\(2\)	\(7\)	\(-4\)
	\(2\)	\(7\)	\(-4\)	\(0\)

The quotient here is \(2x^2+7x-4\) which factorises to \((x+4)(2x-1)\) using Nat 5 methods.

So the full factorisation is \((x+2)(x-1)(x+4)(2x-1).\)

(a)  A factor of \(x-3\) corresponds to a root of \(3,\) so we can use synthetic division like this, taking care to include the \(0\) coefficient of \(x^2.\)

\(3\)	\(2\)	\(-3\)	\(0\)	\(-19\)	\(-24\)
	\(\downarrow\)	\(6\)	\(9\)	\(27\)	\(24\)
	\(2\)	\(3\)	\(9\)	\(8\)	\(0\)

The remainder is \(0\) so \(x-3\) is a factor.

(b)  Now we consider the quotient \(2x^3+3x^2+9x+8.\) As the coefficients are all positive, they clearly don't add to \(0,\) so \(x=1\) isn't a root.

However, if we test \(x=-1\), the odd powers become negative, so \(-2+3-9+8=0\) and we have found our next factor: \(x+1.\) So we need to divide again:

\(-1\)	\(2\)	\(3\)	\(9\)	\(8\)
	\(\downarrow\)	\(-2\)	\(-1\)	\(-8\)
	\(2\)	\(1\)	\(8\)	\(0\)

The quotient here is \(2x^2+x+8\) which does not factorise any further, because the discriminant \(b^2-4ac=-63 \lt 0\). An irreducible quadratic like this appeared on the 2019 Paper 2 and had some candidates criticising the SQA fiercely on social media after the exam. However, it was a perfectly fair question.

The full factorisation is \((x-3)(x+1)(2x^2+x+8).\)

(a)  A factor of \(x-3\) corresponds to a root of \(3,\) so we can use synthetic division like this:

\(3\)	\(1\)	\(-1\)	\(m\)	\(n\)
	\(\downarrow\)	\(3\)	\(6\)	\(3m+18\)
	\(1\)	\(2\)	\(m+6\)	\(3m+n+18\)

The remainder is \(0\) so we obtain the first of two equations: \(3m+n+18=0.\)

Now we turn our attention to the fact that the remainder upon division by \(x-5\) is \(54\).

\(5\)	\(1\)	\(-1\)	\(m\)	\(n\)
	\(\downarrow\)	\(5\)	\(20\)	\(5m+100\)
	\(1\)	\(4\)	\(m+20\)	\(5m+n+100\)

The remainder is \(54\) so we obtain the second of our two equations: \(5m+n+100=54.\)

Now we solve the two equations simultaneously:

$$ \begin{eqnarray} &&3m &+& n &=& -18&&\:\:①\\[6pt] &&5m &+& n &=& -46&&\:\:② \end{eqnarray} $$

$$ \begin{eqnarray} ②-①:\:\:&& 2m &=& -28&&\\[6pt] && \:m &=& -14&& \end{eqnarray} $$

$$ \begin{align} \small\textsf{Subs }\normalsize①:\:\:3m+n &= -18\\[6pt] 3(-14)+n &= -18\\[6pt] -42+n &= -18\\[6pt] n &= -18+42\\[6pt] n &= 24 \end{align} $$

(b)  Now that we know the values of \(m\) and \(n\), finding the other factors and solving the equation is easy.

The first synthetic division in part (a) gave the quotient \(x^2+2x+(m+6),\) which we now know to be \(x^2+2x-8.\)

So we solve the equation like this:

$$ \begin{eqnarray} x^3-x^2+mx+n &=& 0 \\[8pt] x^3-x^2-14x+24 &=& 0 \\[8pt] (x-3)(x^2+2x-8) &=& 0 \\[8pt] (x-3)(x-2)(x+4) &=& 0 \\[8pt] \end{eqnarray} $$

$$ x-3=0\:\:\small\textsf{or}\normalsize\:\:x-2=0\:\:\small\textsf{or}\normalsize\:\:x+4=0 $$ $$ x=3\:\:\small\textsf{or}\normalsize\:\:x=2\:\:\small\textsf{or}\normalsize\:\:x=-4 $$

We can obtain each remainder using synthetic division, like this:

\(-2\)	\(1\)	\(-6\)	\(2\)	\(-p\)
	\(\downarrow\)	\(-2\)	\(16\)	\(-36\)
	\(1\)	\(-8\)	\(18\)	\(-p-36\)

And for the other polynomial:

\(-2\)	\(1\)	\(5\)	\(2p+1\)	\(-37\)
	\(\downarrow\)	\(-2\)	\(-6\)	\(-4p+10\)
	\(1\)	\(3\)	\(2p-5\)	\(-4p-27\)

As the two remainders are equal:

$$ \begin{eqnarray} -p-36 &=& -\!4p-27 \\[8pt] -p+4p &=& -\!27+36 \\[8pt] 3p &=& 9 \\[8pt] p &=& 3 \\[8pt] \end{eqnarray} $$

We will check factors of \(12\):
\(\pm 1,\) \(\pm 2,\) \(\pm 3,\) \(\pm 4,\) \(\pm 6,\) \(\pm 12.\)

Try \(x=1\):  \(1+3-4-12\neq 0.\)

Try \(x=-1\):  \(-1+3+4-12\neq 0.\)

Try \(x=2\):

\(2\)	\(1\)	\(3\)	\(-4\)	\(-12\)
	\(\downarrow\)	\(2\)	\(10\)	\(12\)
	\(1\)	\(5\)	\(6\)	\(0\)

The remainder is \(0\) so \(x-2\) is a factor.

$$ \begin{eqnarray} x^3+3x^2-4x-12 &=& 0 \\[8pt] (x-2)(x^2+5x+6) &=& 0 \\[8pt] (x-2)(x+2)(x+3) &=& 0 \\[8pt] \end{eqnarray} $$

$$ x-2=0\:\:\small\textsf{or}\normalsize\:\:x+2=0\:\:\small\textsf{or}\normalsize\:\:x+3=0 $$ $$ x=2\:\:\small\textsf{or}\normalsize\:\:x=-2\:\:\small\textsf{or}\normalsize\:\:x=-3 $$

We will check factors of \(24\):
\(\pm 1,\) \(\pm 2,\) \(\pm 3,\) \(\pm 4,\) \(\pm 6,\) \(\pm 12,\) \(\pm 24.\)

Try \(x=1\):  \(1-1-10+4+24\neq 0.\)

Try \(x=-1\):  \(1+1-10-4+24\neq 0.\)

Try \(x=2\):

\(2\)	\(1\)	\(-1\)	\(-10\)	\(4\)	\(24\)
	\(\downarrow\)	\(2\)	\(2\)	\(-16\)	\(-24\)
	\(1\)	\(1\)	\(-8\)	\(-12\)	\(0\)

The remainder is \(0\) so \(x-2\) is a factor.

Now we continue the process of finding factors, using the quotient \(x^3+x^2-8x-12.\)

There is no point in trying \(1\) or \(-1\) again, but it is possible that \(2\) is a repeated root, so:

\(2\)	\(1\)	\(1\)	\(-8\)	\(-12\)
	\(\downarrow\)	\(2\)	\(6\)	\(-4\)
	\(1\)	\(3\)	\(2\)	\(-16\)

The remainder \(\neq 0\) so \(x-2\) is not a repeated factor. We will need to keep trying. Let's test \(x=-2\) next:

\(-2\)	\(1\)	\(1\)	\(-8\)	\(-12\)
	\(\downarrow\)	\(-2\)	\(2\)	\(12\)
	\(1\)	\(-1\)	\(-6\)	\(0\)

We're in luck; the remainder was \(0\). So \(x+2\) is another factor. We are now in a position to solve the equation.

$$ \begin{eqnarray} x^4-x^3-10x^2+4x+24 &=& 0 \\[8pt] (x-2)(x+2)(x^2-x-6) &=& 0 \\[8pt] (x-2)(x+2)(x+2)(x-3) &=& 0 \\[8pt] (x-2)(x+2)^{2}(x-3) &=& 0 \\[8pt] \end{eqnarray} $$

$$ x-2=0\:\:\small\textsf{or}\normalsize\:\:x+2=0\:\:\small\textsf{or}\normalsize\:\:x-3=0 $$ $$ x=2\:\:\small\textsf{or}\normalsize\:\:x=-2\:\:\small\textsf{or}\normalsize\:\:x=3 $$

\(a\) and \(b\) are roots of \(f\), so one of them is \(-2\) and the other is \(3\). The factor \(x-b\) is squared, and a turning point on the \(x\)-axis indicates a repeated root, so \(b=3\) and \(a=-2.\)

To find \(k\), we simply substitute the fact that \(f(1)=48\) into what we know of the formula so far:

$$ \begin{eqnarray} f(x) &=& k(x-a)(x-b)^2 \\[6pt] f(x) &=& k(x+2)(x-3)^2 \\[6pt] 48 &=& k(1+2)(1-3)^2 \\[6pt] 48 &=& k(3)(-2)^2 \\[6pt] 48 &=& k(3)(4) \\[6pt] 48 &=& 12k \\[6pt] k &=& 4 \end{eqnarray} $$

Although it is not the only method, you are expected to use the discriminant for questions of this type.

For equal roots:

$$ \begin{eqnarray} b^2-4ac &=& 0 \\[6pt] (k+3)^2-4(1)(4) &=& 0 \\[6pt] (k^2+6k+9)-16 &=& 0 \\[6pt] k^2+6k-7 &=& 0 \\[6pt] (k-1)(k+7) &=& 0 \\[6pt] k-1=0\:\: &\small\textsf{or}\normalsize& \:\:k+7=0 \\[6pt] k=1\:\: &\small\textsf{or}\normalsize& \:\:k=-7 \end{eqnarray} $$

For no real roots:

$$ \begin{eqnarray} b^2-4ac &\lt& 0 \\[6pt] 5^2-4(2)(p+1) &\lt& 0 \\[6pt] 25-8(p+1) &\lt& 0 \\[6pt] 25-8p-8 &\lt& 0 \\[6pt] 17-8p &\lt& 0 \\[6pt] -8p &\lt& -\!17 \\[6pt] p &\gt& \small\frac{17}{8} \end{eqnarray} $$

For two distinct real roots:

$$ \begin{eqnarray} b^2-4ac &\gt& 0 \\[6pt] (-6)^2-4(1)(a) &\gt& 0 \\[6pt] 36-4a &\gt& 0 \\[6pt] -4a &\gt& -36 \\[6pt] a &\lt& 9 \end{eqnarray} $$

Here, \(a=1\small,\) \(b=-n\) and \(c=3\!-\!n\small.\) For two distinct real roots:

$$ \begin{eqnarray} b^2-4ac &\gt& 0 \\[6pt] (-n)^2-4(1)(3-n) &\gt& 0 \\[6pt] n^2-4(3-n) &\gt& 0 \\[6pt] n^2-12+4n &\gt& 0 \\[6pt] n^2+4n-12 &\gt& 0 \\[6pt] (n-2)(n+6) &\gt& 0 \\[6pt] \end{eqnarray} $$

The working above gives us "critical values" of \(n=2\) and \(n=-6\small,\) but in order to state the range of values of \(n\) we need to show evidence that we understand what's going on. The SQA will expect either a table of signs or a quick sketch of \(y=(n-2)(n+6)\small.\)

$$ \begin{array} {|c|c|} \hline n & \rightarrow & -6 & \rightarrow & 2 & \rightarrow \\ \hline y & + & 0 & - & 0 & + \\ \hline \end{array} $$

Alternatively, the graph is a "happy" parabola with a minimum turning point between the roots, so the sections of the graph where \((n-2)(n+6)\gt 0\) are to the left of \(-6\) and to the right of \(2\small.\)

So our solution is: \(n\lt -6\) or \(n\gt 2\small.\)

Area = length\(\times\)breadth = \(x(x-1)\) cm².

So we can create this quadratic inequality:

$$ \begin{eqnarray} x(x-1) &\lt& 30 \\[6pt] x^2-x &\lt& 30 \\[6pt] x^2-x-30 &\lt& 0 \\[6pt] (x+5)(x-6) &\lt& 0 \\[6pt] \end{eqnarray} $$

The working above gives us "critical values" of \(x=-5\) and \(x=6.\)

In order to state the range of values of \(x\) we can consider the graph of \(y=(x+5)(x-6).\)

The graph is a "happy" parabola with a minimum turning point between the roots, so the section of the graph where \((x+5)(x-6)\lt 0\) is between \(-5\) and \(6.\)

So \(-5\lt x\lt 6.\)

But \(x\) represents the length of a rectangle, so negative values make no sense. For this reason, we should restrict our solution to \(0\lt x\lt 6.\)

At National 5, you learned to complete the square on unitary quadratic expressions. Higher takes this one step further. Now the coefficient of \(x^2\) does not have to be \(1.\)

We start by factorising the coefficient of \(x^2\) out of the first two terms, and then calculating the term to add in and immediately subtract out again in exactly the same way as we did at Nat 5.

$$ \begin{eqnarray} && -2x^2+12x+5 \\[6pt] &=& -2(x^2-6x)+5 \\[6pt] &=& -2(x^2-6x+9-9)+5 \\[6pt] &=& -2(x^2-6x+9)-2(-9)+5 \\[6pt] &=& -2(x^2-6x+9)+18+5 \\[6pt] &=& -2(x-3)^2+23 \end{eqnarray} $$

The most common mistake on questions like this is to forget to multiply the \(-9\) by the \(-2\) when breaking it out of the bracket. Take care to follow the steps exactly and you shouldn't go wrong.

This example is going to give us an odd coefficient of \(x\) inside the bracket, which will mean that fractions are unavoidable.

To date, no past paper question has involved fractions, but it would be completely fair of the SQA to set a question like this – so be ready!

$$ \begin{eqnarray} && 4x^2-28x-1 \\[8pt] &=& 4(x^2-7x)-1 \\[8pt] &=& 4\left(x^2-7x+\small\frac{49}{4}\normalsize-\small\frac{49}{4}\normalsize\right)-1 \\[8pt] &=& 4\left(x^2-7x+\small\frac{49}{4}\normalsize\right)-4\left(\small\frac{49}{4}\normalsize\right)-1 \\[8pt] &=& 4\left(x^2-7x+\small\frac{49}{4}\normalsize\right)-49-1 \\[8pt] &=& 4\left(x-\small\frac{7}{2}\normalsize\right)^2-50 \end{eqnarray} $$

At any points of intersection:

$$ \begin{eqnarray} x^2+3x-7 &=& 4x-1 \\[6pt] x^2-x-6 &=& 0 \\[6pt] (x-3)(x+2) &=& 0 \\[6pt] x-3=0 &\:\ \small\textsf{or}\normalsize\ & x+2=0 \\[6pt] x=3 &\:\ \small\textsf{or}\normalsize\ & x=-2 \\[6pt] \end{eqnarray} $$

When \(x=3,\:y=4(3)-1=11\)

When \(x=-2,\:y=4(-2)-1=-9\)

So there are two points of intersection: \(\left(3,\,11\right)\) and \(\left(-\!2,-\!9\right).\)

At the points of intersection:

$$ \begin{gather} x^3-8x+9 = 5x-3 \\[4pt] x^3-13x+12 = 0 \\[6pt] \end{gather} $$

Because \(x=3\) is a root of this equation, we know that \(\left(x-3\right)\) is a factor of \(x^3-13x+12.\)

So we can use synthetic division to factorise \(x^3-13x+12\) fully.

\(3\)	\(1\)	\(0\)	\(-13\)	\(12\)
	\(\downarrow\)	\(3\)	\(9\)	\(-12\)
	\(1\)	\(3\)	\(-4\)	\(0\)

$$ \begin{eqnarray} x^3-13x+12 &=& 0 \\[8pt] (x-3)(x^2+3x-4) &=& 0 \\[8pt] (x-3)(x-1)(x+4) &=& 0 \\[8pt] \end{eqnarray} $$

$$ x-3=0\:\:\small\textsf{or}\normalsize\:\:x-1=0\:\:\small\textsf{or}\normalsize\:\:x+4=0 $$ $$ x=3\:\:\small\textsf{or}\normalsize\:\:x=1\:\:\small\textsf{or}\normalsize\:\:x=-4 $$

When \(x=1,\:y=5(1)-3=2\)

When \(x=-4,\:y=5(-4)-3=-23\)

So the other two points of intersection are \(\left(1,\,2\right)\) and \(\left(-\!4,-\!23\right).\)

Although it isn't the only method, you are expected to use discriminant for questions of this type.

For equal roots:

$$ \begin{eqnarray} b^2-4ac &=& 0 \\[6pt] (3p-2)^2-4(2)(p) &=& 0 \\[6pt] (9p^2-12p+4)-8p &=& 0 \\[6pt] 9p^2-20p+4 &=& 0 \\[6pt] (9p-2)(p-2) &=& 0 \\[6pt] 9p-2=0\:\: &\small\textsf{or}\normalsize& \:\:p-2=0 \\[6pt] p=\small\frac{2}{9}\normalsize\:\: &\small\textsf{or}\normalsize& \:\:p=2 \end{eqnarray} $$

(a)  A factor of \((x+5)\) corresponds to a root of \(-5,\) so we can use synthetic division like this:

\(-5\)	\(1\)	\(3\)	\(-7\)	\(9\)	\(-30\)
	\(\downarrow\)	\(-5\)	\(10\)	\(-15\)	\(30\)
	\(1\)	\(-2\)	\(3\)	\(-6\)	\(0\)

The remainder is \(0\) so \((x+5)\) is a factor.

(b)  Now we consider the quotient \(x^3-2x^2+3x-6.\)

The coefficients \(1-2+3-6\neq 0\) so \(x=1\) isn't a root.

Next we check \(x=-1\) as follows: \(-1-2-3-6\neq 0\small,\) so \(x=-1\) doesn't work either.

The next smallest positive factor of the constant term is \(2\small,\) so we should try that:

\(2\)	\(1\)	\(-2\)	\(3\)	\(-6\)
	\(\downarrow\)	\(2\)	\(0\)	\(6\)
	\(1\)	\(0\)	\(3\)	\(0\)

The quotient here is \((x^2+3)\) which is irreducible; it doesn't factorise further. If you aren't sure about that, check for yourself that its discriminant is negative.

So the full factorisation is \((x+5)(x-2)(x^2+3)\) and we can solve the equation as follows. The irreducible quadratic factor gives us no further real roots.

$$ \begin{eqnarray} x^4+3x^3-7x^2+9x-30 &=& 0 \\[6pt] (x+5)(x-2)(x^2+3) &=& 0 \\[6pt] x+5=0\:\: &\small\textsf{or}\normalsize& \:\:x-2=0 \\[6pt] x=-5\:\: &\small\textsf{or}\normalsize& \:\:x=2 \end{eqnarray} $$