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Algebra
1. Factorising quadratic expressions
We often want to write a quadratic expression as the product of two factors.
E.g. x2 + 5x + 4 = (x + 1)(x + 4)
To factorise a quadratic e.g. x2 + 7x + 6
Step 1: write down two factors of the x2 term
(x )(x )
Step 2: choose factors of the number term (6 in this example) so that they add together to give
the number of xs.
6 = (6)(1)
7=6+1
x2 + 7x + 6 = (x + 1)(x + 6)
but:
2x2 + 5x - 3 = (2x - 1)(x + 3)
3(2) -1 = 5
Difference of two squares
If there are two terms separated by a minus and both terms are perfect squares, then we can
use the difference of two squares formula:
x2 - y2 = (x – y)(x + y)
E.g.
x2 – 25 = (x - 5)(x + 5)
9x2 – 4 = (3x – 2)(3x + 2)
2. Solving quadratic Equations by Factors
If we’re asked to solve: x2 + 7x + 6 = 0
left hand side = the right hand side
we need to find the value(s) of x that make the
We know x2 + 7x + 6 = (x + 1)(x + 6) = 0
So, either
(x + 1) = 0
If (x + 1) = 0
If (x + 6) = 0
equation
or
(x + 6) = 0
then x = -1
then x = -6
If we have factorised 2x2 + 5x - 3 = 0
If (2x + 1) = 0
x = -½
If (x – 5) = 0
x=5
We say -1 and -6 are “the roots” of the
as (2x - 1)(x + 3) = 0
If the equation has fractions to start with, then we have to eliminate the fractions first(using a
common denominator) and then solve like before.
Example:
x 1 1
 1
3
x
common denominator = 3x
x( x  1) 3(1) 3x


     x( x  1)  3( x)  3x    x 2  x  3  3x
3x
3x 3x
x2  2x  3  0
x = -1
or x = 3
(x+1)(x-3)=0
3. Quadratic formula
Sometimes it is hard to factorise using brackets. If so we can use a formula (in the tables) to
solve an expression:
ax2 + bx + c= 0
 b  b 2  4ac
x
2a
E.g Solve correct to 2 decimal places:
a = 1, b = 2, c = -5
the +- means there are 2 solutions
x2 + 2x -5 = 0
 2  22  4(1)( 5)  2  4  (20)  2  24  2  4.898
x



2(1)
2
2
2
+:
x
 2  4.898 2.898

 1.449  1.45
2
2
-:
x
 2  4.898 6.898

 3.449  3.45
2
2
4. Creating Quadratic Equations
If we’re told the roots of an equation, we can work in reverse and create the equation.
We just need to:
(i)
change the sign of the roots
(ii)
put them into brackets after an x
(iii)
multiply the brackets.
Example 1:
form the quadratic equation with roots 2 and -3
(x – 2)(x + 3) = x2 + 3x -2x - 6 = 0
x2 - x - 6 = 0
Example 2:
form the quadratic equation with roots 1/3 and 4
(x – 1/3)(x - 4)
to make multiplication easier we can multiply everything in the first
bracket by 3 (the root is still 1/3).
(3x – 1)(x - 4) = 3x2 - 12x -x + 4 = 0
3x2 - 13x + 4 = 0
5. Simultaneous Equations with squared terms
Important for other topics (e.g. circle)
E.g. solve the following equations: x2 + y2 = 10 and x – y = 4
We can not take one equation away from the other. We must use substitution:
Start with the equation with no squared terms and express one variable in terms of the other.

x=y+4
Substitute this into the other equation so there is now only one variable and the equation can be
solved.
(y + 4)2 + y2 = 10
y2 + 8y + 16 + y2 = 10
2y2 + 8y + 16 - 10 = 0
2y2 + 8y + 6 = 0
y2 + 4y + 3 = 0
(y + 3)(y + 1) = 0
y = -1 or
y = -3
Now for each solution we have to go back and solve for the other variable – it is easiest to go
back to the equation without squared terms.
y = -1: x = y + 4
x = -1 + 4 = 3
y = -3: x = y + 4
x = -3 + 4 = 1
So the solutions are: y = -1, x = 3
and
y = -3, x = 1
6. Algebraic Division
Like long division: e.g.
276 ÷ 12
Example: divide (3x3 + 9x2 + 48) by (x + 4)
Rewrite (3x3 + 9x2 + 48) as (3x3 + 9x2 + 0x + 48)
7. Function notation
A quadratic is an expression with an x2 term
e.g. 9x2 + 2x + 4
A cubic is an expression with an x3 term e.g. 3x2 + 2x2 + x + 1
It’s common practice to use function notation for quadratics and cubics.
E.g.
f(x) = 9x2 + 2x + 4
or
f(x) = 3x3 + 2x2 + 3x + 4
Function notation can be used for substituting values for x.
E.g. if f(x) = x2 - 3x + 8
then
f(4) = 42 – 3(4) + 8 = 16 – 12 + 8 = 12
8. The Factor Theorem
We say that (x-2) is a factor of
Because we can write
f(x) = (x-2)(x+3)
f(x) = x2 + x – 6
However notice that f(2) = 22 + 2 – 6 = 0
Also
f(-3) = (-3)2 + (-3) – 6 = 0
In general the factor theorem says:
1. If (x-a) is a factor of f(x), then f(a) = 0
2. If f(a) = 0, then (x – a) is a factor of f (x)
Example: Show that (x - 2) is a factor of 2x2 - 3x – 2
f(2) must be = 0
Check: 2(2)2 -3(2) – 2 = 8 – 6 - 2 = 0
9. Solving Cubic Equations
x3 - 2x2 - 5x + 6 = 0 is a cubic equation.
If we’re asked to solve this, we must try to values of x that make it true.
We have no formula for this and doing it using brackets would be v. tricky!
If we can find one root of the equation (one value of x that works) then we can reduce the
problem to solving a quadratic equation – then we can use the formula.
Like finding the factors of 154. This is made easier if we know 11 is one factor!
To solve the cubic: we try different values for x in this order: 1, -1, 2, -2 etc.
f(x) = x3 - 2x2 - 5x + 6 = 0
f(1) = (1)3 – 2(1)2 -5(1) + 6 = 0
 1 is a root
 (x-1) is a factor
Now if we divide (x3 - 2x2 - 5x + 6) by (x-1) we can find the other 2 roots.
….. x2 - x - 6
Now factorise this…..(x-3)(x+2)
Factors are: (x-1), (x-3) and (x+2)
Roots are: 1, 3 and -2