**JSS 2 Mathematics Second Term Week 3**

**Topic: FACTORS, FACTORIZATION, FRACTIONS**

**FACTORIZATION OF BINOMIALS**

Binomials are expressions with only two terms being added.

2x^{2} – 4x is an example of a binomial. (You can say that a negative 4x is being added to 2x^{2}.)

First, factor out the GCF, 2x. You’re left with 2x (x – 2). This is as far as this binomial can go. Any binomial in the form 1x +/- n cannot be factored further.

When you have a binomial that is a variable with an even exponent, added to a negative number that has a square root that is a natural number, it’s called a perfect square.

x^{2} – 4 is an example of this. It can be expressed as the product of the square root of the variable plus the square root of the positive constant, and the square root of the variable minus the square root of the positive constant.

Basically, take the square root of the variable. You’ll end up with x. Then square root the 4. You’ll end up with 2. If you add them together, you’ll get x+2. Subtract them, and you’ll get x-2. Multiply the two, and you’ll get (x+4)(x-4). You’ve just factored a perfect square.

If you multiply (x+2)(x-2) together using FOIL, you’ll end back up with x^{2}-4.

(FOIL: First Outer Inner Last, a way of multiplying two binomials together. Multiply the first terms of the binomials (x and x in this case), then the outer two (x and -2), then the inner two (2 and x), then the last terms (2 and -2), then add them all up. x^{2}– 2x + 2x – 4 = x^{^2} – 4.)

This can be done again if one of the binomials is a perfect square, as in this instance:

x^{4} – 16 = (x^{2} + 4) **(x ^{2} – 4)** = (x

^{2}+ 4) (x + 2) (x – 2).

This can be factored further if you bring in irrational numbers, see step [9].

How to factor binomials in the form of (x^{3} **+** b^{3}):

Just plug into (a – b) (a^{2} +ab + b^{2}). For example, (x^{3} + 8) = (x – 2) (x^{2} + 2x + 4).

How to factor binomials in the form of (x^{3} **–** b^{3}):

Plug into (a + b) (a^{2} – ab + b2). Note that the first two signs in the expression are switched.

(x^{3} – 8) = (x + 2) (x^{2} – 2x + 4).

**ALGEBRAIC FRACTIONS**

**Adding and subtracting**

Algebraic fractions are simply fractions with algebraic expressions on the top and/or bottom.

When adding or subtracting algebraic fractions, the first thing to do is to put them onto a common denominator (by cross multiplying).

e.g. 1 + 4

(x + 1) (x + 6)

= 1(x + 6) + 4(x + 1)

(x + 1)(x + 6)

= x + 6 + 4x + 4

(x + 1)(x + 6)

= 5x + 10

(x + 1)(x + 6)

**Solving equations**

When solving equations containing algebraic fractions, first multiply both sides by a number/expression which removes the fractions.

**Example**

Solve 10 -2 = 1

(x + 3) x

multiply both sides by x(x + 3):

∴ 10x(x + 3) – 2x(x + 3) = x(x + 3)

(x + 3) x

∴ 10x – 2(x + 3) = x^{2} + 3x [after cancelling]

∴ 10x – 2x – 6 = x^{2} + 3x

∴ x^{2} – 5x + 6 = 0

∴ (x – 3)(x – 2) = 0

∴ either x = 3 or x = 2

**Multiplication**

To multiply algebraic fractions, factorise the numerators and denominators. Then cancel the factors common to the numerator and denominator before applying multiplication to obtain the answer.

**Example**

Simplify

a. 7 X b/8 b. p/14 X 6/p c. x – 3/8 X 12/x – 3

Solution

a. 7 X b/8 = ab/56

b. p/14 X 6/p = 3/7

c. x – 3/8 X 12/x – 3 = 3/2

**FRACTIONS WITH BRACKETS**

x + 6/3 is a short way of writing (x + 6)/3 or 1/3(x + 6).

Notice that all the terms of the numeratorare divided by 3.

X + 6/3 = (x + 6)/3 = 1/3(x + 6) = 1/3x + 2

**Example**

Simplify

a. x + 3/5 + 4x – 2/5

**Solution**

x + 3/5 + 4x – 2/5

= (x + 3) + (4x – 2)/5

= x + 3 + 4x – 2/5

= 5x + 1/5

**Exercise**

Simplify the following

a. 2a – 3/2 + a + 4/2

b. 3b + 4/3 + 2b – 5/3

c. 4c – 3/5 – 2c + 1/5

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