(a) Show that is a factor of (b) Hence factorise fully.
(a) A factor of corresponds to a root of so we can use synthetic division like this:
The remainder is so is a factor.
(b) It is important to understand that the synthetic division in part (a) divided the cubic by not
Now we consider the quotient It factorises as which further factorises to
So the full factorisation is which is much neater if we multiply the by the first factor and present our final answer as
Example 2 (non-calculator)
Factorise fully.
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 first. Substituting into the cubic, we obtain It would have been quicker here just to add the coefficients. Anyway, we're in luck because it comes out to be So by the factor theorem, is a root and so is a factor.
Now we can use synthetic division to find the quotient after division by
(a) Show that is a factor of (b) Hence factorise fully.
(a) A factor of corresponds to a root of so we can use synthetic division like this, taking care to include the coefficient of
The remainder is so is a factor.
(b) Now we consider the quotient As the coefficients are all positive, they clearly don't add to so isn't a root.
However, if we test , the odd powers become negative, so and we have found our next factor: So we need to divide again:
The quotient here is which does not factorise any further, because the discriminant . 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
Example 5 (non-calculator)
For the polynomial
is a factor
54 is the remainder when it is divided by
(a) Determine the values of and (b) Hence solve
(a) A factor of corresponds to a root of so we can use synthetic division like this:
The remainder is so we obtain the first of two equations:
Now we turn our attention to the fact that the remainder upon division by is .
The remainder is so we obtain the second of our two equations:
Now we solve the two equations simultaneously:
(b) Now that we know the values of and , finding the other factors and solving the equation is easy.
The first synthetic division in part (a) gave the quotient which we now know to be
So we solve the equation like this:
Example 6 (non-calculator)
The same remainder is found when and are divided by Find the value of
We can obtain each remainder using synthetic division, like this:
Find the range of values of for which has two distinct real roots.
Here, and For two distinct real roots:
The working above gives us "critical values" of and but in order to state the range of values of 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
Alternatively, the graph is a "happy" parabola with a minimum turning point between the roots, so the sections of the graph where are to the left of and to the right of
So our solution is: or
Example 14 (non-calculator)
A rectangle has length cm and a breadth that is cm shorter than the length. Its area is less than cm2. Find the range of possible values of
Area = lengthbreadth = cm2.
So we can create this quadratic inequality:
The working above gives us "critical values" of and
In order to state the range of values of we can consider the graph of
The graph is a "happy" parabola with a minimum turning point between the roots, so the section of the graph where is between and
So
But represents the length of a rectangle, so negative values make no sense. For this reason, we should restrict our solution to
Example 15 (non-calculator)
Express in the form
At National 5, you learned to complete the square on unitary quadratic expressions. Higher takes this one step further. Now the coefficient of does not have to be
We start by factorising the coefficient of 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.
The most common mistake on questions like this is to forget to multiply the by the when breaking it out of the bracket. Take care to follow the steps exactly and you shouldn't go wrong.
The equation has equal roots. Determine the possible values of
Although it isn't the only method, you are expected to use discriminant for questions of this type.
For equal roots:
Example 20 (non-calculator)
SQA Higher Maths 2023 Paper 1 Q10
(a) Show that is a factor of (b) Hence, or otherwise, solve
(a) A factor of corresponds to a root of so we can use synthetic division like this:
The remainder is so is a factor.
(b) Now we consider the quotient
The coefficients so isn't a root.
Next we check as follows: so doesn't work either.
The next smallest positive factor of the constant term is so we should try that:
The quotient here is 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 and we can solve the equation as follows. The irreducible quadratic factor gives us no further real roots.
Example 21 (non-calculator)
SQA Higher Maths 2024 Paper 1 Q8
The equation has no real roots. Determine the range of values of Justify your answer.
For no real roots:
Our critical values are therefore and
To obtain the range of values of with suitable justification to receive full marks, we should either sketch the parabola or use a table of values.
Either way, the range of values is
Example 22 (non-calculator)
SQA Higher Maths 2024 Paper 1 Q10
(a) Show that is a factor of (b) Hence, or otherwise, factorise fully.
(a) A factor of corresponds to a root of so we can use synthetic division like this:
The remainder is so is a factor.
(b) Now we consider the quotient
The coefficients so isn't a root.
Next we check as follows: so works and is another factor.
So we divide again, by
The quotient here is which factorises further, to
So the full factorisation is
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