Download Factor and Remainder Theorem Revision 6

Factors of Equations
By Charlie Williams
Two ways of solving a cubic once you have one factor
Method 1: Equating coefficients
This is the method we have done in class and involves
equating coefficients and comparing them to find the
answer.
This method is good because it is easy to learn and it’s
easier to understand what you’re doing, although it is
not as fast and can be easy to make mistakes so you
need to check your work carefully.
There is an example of how to do it on the next slide
Example
Take the equation x³+6x²+11x+6=0 and you have been given or worked out
that (x+1) is a factor.
The first step is you say:
x³+6x²+11x+6 ≡ (x+1)(ax²+bx+c)
You then expand the brackets on the right hand side and simplify which will
give you (the brackets are added to make it clearer):
x³+6x²+11x+6 ≡ ax³+(a+b)x²+(b+c)x+c
This is where you actually compare coefficients by looking at both sides and
seeing what matches up, and you are able to make some easily solvable
linear equations which will give you a, b and c:
a=1
a+b=6
b+c=11
c=6
(continues on next slide)
From this you can quickly work out that a is 1, b is 5 and c is 6.
This means that if you divide the cubic equation x³+6x²+11x+6 by (x+1) you
will be left with the quadratic x²+5x+6, which you write as:
(x+1)(x²+5x+6)
This can quickly be factorised again as it is a quadratic giving:
(x+1)(x+2)(x+3)
And therefore the roots of the equation are -1, -2 and -3.
Sometimes you will not be able to factorise the quadratic, in which case
leaving it in that form is ok.
Algebraic Long Division
This is the other method of solving a cubic
once you have one factor
The advantages of this are it is faster, a lot
neater and you are very unlikely to make
mistake, although it is complicated to
learn.
There are a couple of examples of how to
do it on the next few slides
Example
You are given the equation x³-x²-10x-8 and you know (x-2) is a factor
It is quite complicated so I will show the working then explain it on the next
slides.
(x+2)
x³
-
x²(x+2)
x³
+
-3x(x+2)
x²
-
10x
-
8
2x²
-3x²
-
10x
-3x²
-
6x
-4x
-
8
-4x
-
8
-4(x+2)
Giving you (x+2)(x²-3x-4) which factorises to (x+2)(x+1)(x-4)
So your roots are -2, -1 and 4
The first thing you do is write out the factor on the left and the
cubic on the right, leave it quite spread out to give you room for
working
You then see what you would have
to multiply the x term in the factor
(the thing in the brackets), which is
just here, by to get the first thing in
the cubic, in this case x² as x²*x=x³
Write this outside the bracket and
then expand the bracket which
here gives x³+2x², and write it
under the original cubic.
Now draw a line under it, and
minus 2x² from -x², giving you
-3x² and write it underneath. Also
bring down the x term in the
original cubic, and write it next to
what you’ve minused
Again see what you would need to multiply the first part of
the factor by to get, in this case, -3x², which is -3x
(-3x*x=-3x²) and write it next to the factor and expand the
brackets, giving -3x²-6x
Minus -6x from -10x, and due to
the double negative this gives you
-4x, write it below the line and
bring down the constant (here -8)
and write it next to it.
Look to see what you need to
multiply the x term in the factor by
to get -4x, which is -4, write this
next to the factor and expand the
brackets.
To get your final answer look down the left hand side,
at the numbers outside the brackets which gives you
x²-3x-4, so x³-x²-10x-8 = (x+2)(x²-3x-4) which
factorises to (x+2)(x-4)(x+1), so your roots are -2, 4
and -1
(One good thing about
this method is if these
two don’t match up you
know you’ve gone
wrong, so it’s harder to
make a mistake)
Here’s another example:
You have the cubic x³+2x²-5x-6 and know (x+3) is a factor
(x+3)
x³
+
2x²
x²(x+3)
x³
+
3x²
-x(x+3)
-2(x+3)
-x²
-x²
-
5x
-
-
6
-
6
6
5x
3x
-2x
-2x
These two match up so you know you’re right.
This gives you (x+2x²-5x-6) = (x+3)(x²-x-2) which factorises to
(x+3)(x+1)(x-2) so your roots are 2, -1 and -3.