Higher Maths
Polynomials and Quadratics

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

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 \((x+\tfrac{3}{2})\small,\) not \((2x+3)\small.\)

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

So the full factorisation is \((x+\tfrac{3}{2})(2)(x-1)(x+1),\) which can be simplified by multiplying the \(2\) by the first factor and presenting our final answer in the form \((2x+3)(x-1)(x+1)\small.\)

Video solution by Clelland Maths 

This question doesn't have a part (a) to guide us towards one of the factors, so we have to start from scratch.

Thankfully, there are a few easy tricks to help us find the first factor and use it to complete the factorisation.

The first such trick is: 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 equals \(0\small.\) 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 \((x-1)\small.\)

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

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

Note: Don't make the mistake of assuming that every quadratic factor will factorise into two linear factors. Some quadratics are irreducible, as you saw when learning about the disriminant back in N5. So a cubic polynomial will factorise into either three linear factors or one linear factor times an irreducible quadratic factor.

Video solution by Clelland Maths 

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

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

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

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

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

Video solution by Clelland Maths 

(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 \(x^2\small.\)

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

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

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

The quotient here is \(2x^2+x+8\) which does not factorise any further, because the discriminant \(b^2-4ac=-63 \lt 0\small.\)

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

Note: An irreducible quadratic like this appeared on 2019 Paper 2  and had some candidates fiercely criticising the SQA on social media after the exam. However, it was a perfectly fair question.

Video solution by Clelland Maths 

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

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

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

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

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 \(n\small,\) finding the other factors and solving the equation is easy.

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

So we solve the equation like this:

$$ \begin{gather} x^3-x^2+mx+n=0\\[8pt] x^3-x^2-14x+24=0\\[8pt] (x-3)(x^2+2x-8)=0\\[8pt] (x-3)(x-2)(x+4)=0\\[8pt] x-3=0\:\:\small\textsf{or}\normalsize\:\:x-2=0\:\:\small\textsf{or}\normalsize\:\:x+4=0\\[8pt] x=3\:\:\small\textsf{or}\normalsize\:\:x=2\:\:\small\textsf{or}\normalsize\:\:x=-4 \end{gather} $$

Video solution by Clelland Maths 

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

And for the other polynomial:

As the two remainders are equal:

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

Video solution by Clelland Maths 

We start by checking positive and negative factors of \(12\small.\)

First, try \(x=1\!:\)  \(1+3-4-12\neq 0\small.\)

Then try \(x=-1\!:\)  \(-1+3+4-12\neq 0\small.\)

As they didn't work, try \(x=2\!:\)

The remainder is \(0\) so \((x-2)\) is a factor, and we can complete the solution like this:

$$ \begin{gather} x^3+3x^2-4x-12=0\\[8pt] (x-2)(x^2+5x+6)=0\\[8pt] (x-2)(x+2)(x+3)=0\\[8pt] x-2=0\:\:\small\textsf{or}\normalsize\:\:x+2=0\:\:\small\textsf{or}\normalsize\:\:x+3=0\\[8pt] x=2\:\:\small\textsf{or}\normalsize\:\:x=-2\:\:\small\textsf{or}\normalsize\:\:x=-3\\[8pt] \end{gather} $$

Video solution by Clelland Maths 

We start by checking positive and negative factors of \(24\small.\)

First try \(x=1\!:\)  \(1-1-10+4+24\neq 0.\)

Then try \(x=-1\!:\)  \(1+1-10-4+24\neq 0.\)

As they didn't work, try \(x=2\):

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

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

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

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

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

$$ \begin{gather} x^4-x^3-10x^2+4x+24=0\\[8pt] (x-2)(x+2)(x^2-x-6)=0\\[8pt] (x-2)(x+2)(x+2)(x-3)=0\\[8pt] (x-2)(x+2)^{2}(x-3)=0\\[8pt] x-2=0\:\:\small\textsf{or}\normalsize\:\:x+2=0\:\:\small\textsf{or}\normalsize\:\:x-3=0\\[8pt] x=2\:\:\small\textsf{or}\normalsize\:\:x=-2\:\:\small\textsf{or}\normalsize\:\:x=3 \end{gather} $$

Video solution by Clelland Maths 

\(a\) and \(b\) are roots of \(f\small,\) so one of them is \(-2\) and the other is \(3\small.\)

The factor \((x-b)\) is squared, and a turning point on the \(x\)-axis indicates a repeated root, so \(b=3\) and \(a=-2\small.\)

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

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

Video solution by Clelland Maths 

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

For equal roots, the discriminant equals zero.

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

Video solution by Clelland Maths 

For no real roots, the discriminant  is negative.

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

Video solution by Clelland Maths 

For two distinct real roots, the discriminant  is positive.

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

Video solution by Clelland Maths 

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

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

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. Qualifications Scotland expects 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 \(n\gt 2\small.\)

Video solution by Clelland Maths 

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

So we can create this quadratic inequality:

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

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

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

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 \(6\small.\)

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

However, that isn't our final answer. We need to refer back to the question and think about the physical context.

\(x\) represents the length of a rectangle, so negative values make no sense. Also, as the breadth is \(1\) cm shorter than the length, the length cannot be \(1\) cm or less.

For these reasons, we should restrict our solution to \(1\lt x\lt 6\small.\)

Video solution by Clelland Maths 

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

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. If you follow the above steps exactly, you shouldn't go wrong.

Note: The working shown above is longer than it needs to be, so that the process may be clearly understood. If you have been taught a shorter form of working, just use that.

Video solution by Clelland Maths 

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 Qualifications Scotland to set a question like this – so be ready!

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

Video solution by Clelland Maths 

At any points of intersection:

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

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 \(\left(-2,-9\right)\small.\)

Video solution by Clelland Maths 

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 \(x^3-13x+12\small.\)

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

Now we can finish solving the equation:

$$ \begin{gather} x^3-13x+12=0\\[8pt] (x-3)(x^2+3x-4=0\\[8pt] (x-3)(x-1)(x+4)=0\\[8pt] x-3=0\:\:\small\textsf{or}\normalsize\:\:x-1=0\:\:\small\textsf{or}\normalsize\:\:x+4=0\\[8pt] x=3\:\:\small\textsf{or}\normalsize\:\:x=1\:\:\small\textsf{or}\normalsize\:\:x=-4 \end{gather} $$

To finish the question, we find the matching \(y\)-values and state the coordinates.

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 \(\left(-4,-23\right).\)

Video solution by Clelland Maths 

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

For equal roots, the discriminant equals zero.

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

Video solution by Clelland Maths 

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

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

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

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:

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.

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

Video solution by Clelland Maths 

For no real roots, the discriminant  is negative.

$$ \begin{gather} b^2-4ac\lt 0 \\[6pt] (m-4)^2-4(1)(2m-3)\lt 0 \\[6pt] m^2-8m+16-4(2m-3)\lt 0 \\[6pt] m^2-8m+16-8m+12\lt 0 \\[6pt] m^2-16m+28\lt 0 \\[6pt] (m-2)(m-14)\lt 0 \\[6pt] \end{gather} $$

Our 'critical values' are therefore \(2\) and \(14\small.\)

To obtain the range of values of \(m\small,\) with suitable justification to receive full marks, we should either sketch the parabola \(y=(m-2)(m-14)\) or use a table of values.

Either way, the range of values is \(2\lt m \lt 14\small.\)

Video solution by Clelland Maths 

(a)  A factor of \((x-1)\) corresponds to a root of \(1\small,\) so we can use synthetic division like this:

The remainder is \(0\) so \((x-1)\) is a factor.

(b)  Now we consider the quotient \(2x^3+5x^2+x-2\small.\)

The coefficients \(2+5+1-2\neq 0\) so \(x=1\) isn't a root.

Next we check \(x=-1\) as follows: \(-2+5-1-2=0\small,\) so \(x=-1\) works and \(x+1\) is another factor.

So we divide again, by \((x+1)\small.\)

The quotient here is \((2x^2+3x-2)\) which factorises further, to \((x+2)(2x-1)\small.\)

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

Video solution by Clelland Maths 

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

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

(b)  We solve the equation by fully factorising the cubic expression.

$$ \begin{gather} 5x^3+16x^2-x-12=0\\[7pt] (x+3)(5x^2+x-4)=0\\[7pt] (x+3)(5x-4)(x+1)=0\\[7pt] x+3=0\:\:\small\textsf{or}\normalsize\:\:5x-4=0\:\:\small\textsf{or}\normalsize\:\:x+1=0\\[7pt] x=-3\:\:\small\textsf{or}\normalsize\:\:x=\small\frac{4}{5}\normalsize\:\:\small\textsf{or}\normalsize\:\: x=-1\\[7pt] \end{gather} $$

Video solution by Clelland Maths 

For two distinct real roots, the discriminant  is positive. So:

$$ \begin{gather} b^2-4ac\gt 0\\[6pt] (3k)^2-4(9)(k)\gt 0\\[6pt] 9k^2-36k\gt 0\\[6pt] k^2-4k\gt 0\\[6pt] k(k-4)\gt 0\\[6pt] \end{gather} $$

So our 'critical values' are \(k=0\) and \(k=4\small.\)

To determine the range of values for \(k\small,\) we could either make a table of signs (as in example 13 above) or draw a quick sketch of \(y=k(k-4)\small.\)

The parts of the graph that are above the horizontal axis, i.e. for which \(y\gt 0\small,\) are our solutions.

So, for two distinct real roots, \(k\lt 0\) or \(k \gt 4\small.\)

Video solution by Clelland Maths 

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.

$$ \begin{eqnarray} && \!\!\!\!\!\!2x^2+16x+5 \\[6pt] &=& 2(x^2+8x)+5 \\[6pt] &=& 2(x^2+8x+16-16)+5 \\[6pt] &=& 2(x^2+8x+16)-2(16)+5 \\[6pt] &=& 2(x+4)^2-32+5 \\[6pt] &=& 2(x+4)^2-27 \end{eqnarray} $$

Note: There are various other ways of laying out the working for completion of the square. Your working doesn't need to be quite as detailed as that shown above.

Video solution by Clelland Maths