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Revision:OCR Core 2 - Factors and remainders

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 2 - Factors and remainders

1 8. Factors and remainders
2 Also See

8. Factors and remainders

Factorising quadratics

Although in many cases it is rather easy to factorise quadratics, it is not always so easy. As the coefficients become larger, the process can become very difficult, and therefore it is desirable to have some form of method to the factorising process by which one can be sure that one will not be wasting time by checking factors which one does not necessarily need to check.

Consider the idea that one has a quadratic:

p(x) = (nx + a)(mx + b)

Consider $x = \frac{-a}{n}$ .

In this case:

$\displaystyle p \left( \frac{-a}{n} \right) = 0$

Also, one can consider:

$x = \frac{-b}{m}$

$\displaystyle p \left( \frac{-b}{m} \right) = 0$

This has significance as one can say that:

For $ax^{2} + bx + c$ , if $\displaystyle p \left( \frac{-a}{n} \right) = 0$ , (nx + a) is a factor.

This is called the extended factor theorem for quadratics.

There is a more specific, and simple version of this theorem:

Consider p(x) = (x - a)(x - b)

p(a) = 0

p(b) = 0

For $ax^{2} + bx + c$ , if p(a) = 0 , (x - a) is a factor.

This is the factor theorem for quadratics.

Both of these ideas can be used in order to speed up the process of factorising a quadratic.

After one has a single factor, one can find the other, as one knows that the constant value of the quadratic is formed by the multiplication of the two constant values of the factors, and also that the coefficient of the $x^{2}$ term is formed by the multiplication of the coefficients of the "x" terms in the factors. One can deduce the second factor from these observations.

Equating coefficients

If one is aware of one factor, one can use a method called equating coefficients to find the other, corresponding factor.

This method is often useful in situations where one is dealing with higher-degree polynomials, but the same idea applies to all polynomials.

Consider:

$p(x) = ax^{2} + bx + c$

If p(d) = 0 , (x - d) is a factor, and therefore one knows that there must be some factor such that:

$(x - d)(Ax + B) \equiv p(x)$

Hence:

$Ax^{2} + (B - d)x - dB \equiv p(x)$

From this one can calculate the values of "A", and "B", and therefore the other factor.

This method can be applied whenever one knows one of the factors.

One might notice that the identity symbol is used, this is because the two sides are equal for all values of "x".

One can prove that this system works.

$ax^{2} + bx + c \equiv dx^{2} + ex + f$ .

$p(x) = ax^{2} + bx + c$

p(0) = c

p(1) = a + b +c

p(-1) = a - b + c

$q(x) = dx^{2} + ex + f$

q(0) = f

p(0) = q(0) (Due to the identity)

Therefore:

c = f

q(1) = d + e + f

p(1) = q(1) (Due to the identity)

d + e + f = a + b + c

Hence:

d + e = a + b

Also:

p(-1) = q(-1) (Due to the identity)

a - b + c = d - e + f

Hence:

a - b = d - e

Now (summing these simultaneous equations):

2a = 2d

a = d

b = e

Hence the equation of coefficients is correct.

Factorising cubics

Often one will need to factorise a cubic (many times this will be in order to find the x-axis intercepts of a cubic function, or the stationary points of quartic function).

All the methods which applied with the quadratics will apply with the cubics.

Example

1. Factorise the following:

$x^{3} + 3x^{2} + 3x + 1$ .

One merely states that:

$p(x) = x^{3} + 3x^{2} + 3x + 1$

p(-1) = 0

Hence (x + 1) is a factor.

Hence:

$(x + 1)(Ax^{2} + Bx + C) \equiv p(x)$

A = 1, B = 2, C = 1

Hence:

$(x + 1)(x^{2} + 2x + 1) \equiv p(x)$

Now one can attempt to factorise the quadratic:

$(x^{2} + 2x + 1) = q(x)$

p(-1) = 0

Hence (x + 1) is a factor.

$(x + 1)(Dx + E) \equiv q(x)$

D = 1, E = 1

Hence:

$(x + 1)(x + 1)(x + 1) \equiv p(x)$

Division of polynomials

One will have come across the concept of a remainder regarding the division of constants, however the same idea applies to polynomials.

If one was to attempt to equate the coefficients with an expression that was not a factor of the polynomial, one would obtain inconsistent results.

For example:

$p(x) = x^{2} + 2x + 1$

p(1) = 1 + 2 + 1 = 4

Therefore (x - 1) is not a factor.

$(x - 1)(Ax + B) \equiv p(x)$

$Ax^{2} + (B - A)x - B \equiv p(x)$

A = 1, B = -1, but that means that (B - A)x = -2x , this is the inconsistency.

One needs to bring in the idea of a remainder.

Instead of equating the coefficients in the usual manner, add a variable for the remainder.

$(x - 1)(Ax + B) + R \equiv p(x)$

R - B = 1

B - A = 2

A = 1

B = 1

R = 4

Therefore:

$p(x) \equiv (x - 1)(x + 1) + 4$ .

(One might notice that the remainder is equal to the value of p(1) , this will be discussed later in these notes).

One can say that when one has a polynomial divided by a linear polynomial, one will be left with another polynomial (the quotient) and a remainder (which might be 0).

Hence:

$p(x) \equiv (ax + b)q(x) + R$

As one is dividing by a linear polynomial, q(x) will have a degree which is one less than that of p(x)

The remainder theorem

As one has seen, if $\displaystyle p \left( \frac{a}{b} \right) = 0$ , (bx - a) is a factor of p(x) .

It is now useful to consider the significance of this figure during the process of dividing two polynomials, and taking the remainder.

Take the general form of the resultant expression after a division by a linear polynomial:

$p(x) \equiv (ax + b)q(x) + R$ .

As this is an expression of identical equality, one can choose any number for x, and the equality is preserved.

One wishes to isolate the "R" term in this explanation, and therefore it would be useful for either q(x) = 0 , and/or (ax + b) = 0 . As one is aware of the value of it is easy to ensure that this is 0.