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ALGEBRAIC METHODS
You can simplify algebraic fractions using division.
โ When simplifying an algebraic fraction, where possible, factorise the numerator and
denominator.
โ Then, you can cancel out the common factors resulting in a simplified expression.
Note : you can freely cancel common factors in an expression but not when it is in the form
of an equation. For instance, take ๐(๐ฅ) = 0 assuming that ๐(๐ฅ)is a polynomial and not
linear. We cannot cancel out ๐ฅinstead but instead we factorise this variable out, such that
we do not lose solutions where ๐ฅ = 0 (as in the case that we divided by ๐ฅwe would have to
impose the condition that ๐ฅ โ 0since otherwise the function would be undefined.
Polynomials are finite expressions with positive whole number indices.
โ You can use long division to divide a polynomial by (๐ฅ ยฑ ๐), where ๐is a constant.
โ The order of a polynomial can be defined by the highest power of ๐ฅ, or any variable that
changes
with respect to ๐(๐ฅ). For instance, say some function ๐ (๐ก) = 3๐ก - 9 then ๐ (๐ก)can be
referred to as
a degree one polynomial.
The factor theorem can be used to find simple linear factors of a polynomial.
โ The factor theorem states that if ๐(๐ฅ)is a polynomial thenโฆ
- If ๐(๐) = 0 then (๐ฅ โ ๐) is a factor of ๐(๐ฅ)
- Conversely, if (๐ฅ โ ๐) is a factor of ๐(๐ฅ) then ๐(๐) = 0
โ When provided an expression, we can change its form to become a function instead
which will
allow us to make use of the appropriate notation when substituting in values for ๐ฅto
prove that
๐(๐ฅ) takes a certain value, which are usually the roots of the function at ๐(๐ฅ) = 0. In
other words,
it is much simpler to substitute ๐ฅ = 2 by saying ๐(2)which only makes sense when we
know what
๐(๐ฅ) represents. Without this notation, we would have to write let ๐ฅ = 2 โ๐ฆ = ...
For instance, take the equation of ๐ฆ = 2๐ฅยณ โ 8๐ฅยฒ + ๐ฅ We can let ๐ฆ = ๐(๐ฅ) so that ๐(๐ฅ) = 2๐ฅยณ โ 8๐ฅยฒ + ๐ฅ
โ In the case that we are provided with a polynomial and asked to factorise it, we can
simply
substitute some probable solutions of ๐ฅinto the function. Say we are trying to find the
values of
๐ฅ for which ๐(๐ฅ) = 0 and initially try to substitute ๐ฅ = ยฑ๐. Considering the case that for
this
particular function ๐(๐) = 0; by the factor theorem, we know that (๐ฅ โ ๐)is a factor of
๐(๐ฅ).
โ Ultimately, we are just using trial-and-error to find one solution of the equation and the
factor of
the polynomial that is derived from this is what we are dividing the dividend by (the
divisor).
โ In an ideal situation we would not have to use polynomial division and we would be able
to
factorise to find and cancel common factors. However, this method is especially
effective for
larger polynomials. We could also do polynomial division by inspection which involves
the
comparison of the coefficients each descending power of ๐ฅfrom its highest index.
Remember: always show your full workings when substituting values of ๐ฅinto a function.
Mathematical proofs must showcase that a statement is true in every case.
โ Mathematical statements that are yet to be proven are known as conjectures.
โ Mathematical statements that have been proven are called theorems.
Below is an example of a proof, which in this case has been done by deduction.
Statement: the product of two odd numbers is always odd.
E xample 2
Prove that if (๐ฅ โ ๐) is a factor of ๐(๐ฅ) then ๐(๐) = 0
If (๐ฅ โ ๐) is a factor of ๐(๐ฅ) then
๐(๐ฅ) = (๐ฅ โ ๐) ร ๐(๐ฅ)
โด ๐(๐) = (๐ โ ๐) ร ๐(๐)
i.e. ๐(๐) = 0 ร ๐(๐)
โด ๐(๐)= 0, as required
This proof even holds when ๐(๐ฅ) is a degree one polynomial as ๐(๐ฅ) can take any form.
E xample 3
Prove that ๐ด(1, 1), ๐ต(3, 3), ๐ถ(4, 2) are the vertices of a right-angled triangle
The gradient of line ๐ด๐ต = 3โ 3โ1 =^
2 2 = 1
The gradient of line ๐ต๐ถ = 2โ 4โ3 =^
โ 1 =^ โ^1
The gradient of line ๐ด๐ถ =
2โ 4โ1 =^
1 3
The gradients are different so the points are not collinear.
Hence ๐ด๐ต๐ถis a triangle.
Gradient of ๐ด๐ต ร gradient of๐ต๐ถ = 1 ร (โ 1) = โ 1
โด ๐ด๐ต is perpendicular to ๐ต๐ถ, and
the triangle is a right-angled triangle.
For proofs involving proving geometric results, it can be helpful to sketch a diagram.
E xample 4
The equation ๐๐ฅยฒ + 3๐๐ฅ + 2 = 0 , where ๐ is a constant that has no real roots
Prove that ๐ satisfies the inequality 0 โค ๐ โค (^89)
๐๐ฅยฒ + 3๐๐ฅ + 2 = 0 has no real roots
โด ๐ยฒ โ 4๐๐ < 0
When ๐ = 0 ,
(0)๐ฅยฒ + 3(0)๐ฅ + 2 = 0
2 = 0
Which is impossible, so no real roots whenโฆ
๐ = 0 and 0 < ๐ < (^89)
Combining the range of values of ๐for which the original quadratic has no real roots:
0 โค ๐ โค 89 as required
We cannot substitute ๐ = 0 into the discriminant so we substitute it into the original
expression
You need to be able to use three types of proofs to prove or disprove mathematical
statements.
โ Proof by deduction
This proof requires you to start from known facts and end up at the conclusion using
logical
and deductive steps. It is not acceptable to start with the conclusion and then verify that it
works because you are assuming the statement that you are trying to prove.
โ Proof by exhaustion
This proof involves breaking the statement into smaller cases and proving each case
separately.
This may require you to split up the cases into odd and even or positive and negative
numbers.
Proof by exhaustion is more well-suited when you only need to determine a small number
of results.
You cannot use one example to prove that a statement is true, as one example is only one
case, so there is no element of comparison involved.
๐ฅ๐ฆ โฅ^2
Now to start the proof, letโs consider (๐ฅ โ ๐ฆ)ยฒ
(๐ฅ โ ๐ฆ)ยฒ โฅ 0
๐ฅยฒ + ๐ฆยฒ โ 2๐ฅ๐ฆ โฅ 0
๐ฅยฒ + ๐ฆยฒ โ 2๐ฅ๐ฆ ๐ฅ๐ฆ
This step is valid because ๐ฅ and ๐ฆ are positive so๐ฅ๐ฆ > 0
๐ฅยฒ ๐ฅ๐ฆ +^
๐ฅ๐ฆ โ^
๐ฅ๐ฆ โฅ^0
๐ฅ +^
๐ฆ โ^2 โฅ^0
๐ฆ +^
๐ฅ โฅ^2
