Simplifying Algebraic Fractions

Divide the top and the bottom by the highest common factor of 8x and 12x^2.

1. The highest common factor of 8x and 12x^2 is 4x.
2. 8x\div4x=2 and 12x^2\div4x=3x.
3. So \frac{8x}{12x^2}=\frac{2}{3x}.

Take out a common factor in the numerator.

1. Factorise the numerator: x^2+5x=x(x+5).
2. So the fraction is \frac{x(x+5)}{x}. Cancel the common factor x.
3. The result is x+5.

(x-2)^2 means (x-2)(x-2), so (x-2) is a factor of both the top and the bottom.

1. Write the denominator as 9(x-2)(x-2). One factor of (x-2) is common to the top and the bottom, so it cancels, leaving \frac{6}{9(x-2)}.
2. Divide the 6 and the 9 by 3: \frac{2}{3(x-2)}.

The numerator is a difference of two squares.

1. Use the difference of two squares: a^2-16=a^2-4^2=(a-4)(a+4).
2. Cancel the common factor a-4.
3. The result is a+4.

Factorise the numerator; one of its factors should be x+2.

1. Factorise the numerator: two numbers that multiply to 10 and add to 7 are 2 and 5, so x^2+7x+10=(x+2)(x+5).
2. Cancel the common factor x+2.
3. The result is x+5.

The top is a difference of two squares and the bottom is a perfect square.

1. Factorise the numerator as a difference of two squares: x^2-9=(x-3)(x+3).
2. The denominator is a perfect square: x^2+6x+9=(x+3)(x+3).
3. Cancel one factor of x+3 to obtain \frac{x-3}{x+3}.

Take out the highest common factor of the numerator first.

1. Factorise the numerator: 3x(x+4).
2. Cancel 3x against 6x, leaving a denominator of 2.
3. The result is \frac{x+4}{2}.

Factorise the numerator; one of its factors should be x-3.

1. To factorise 2x^2-5x-3, find two numbers that multiply to 2\times(-3)=-6 and add to -5: they are -6 and 1.
2. Split the middle term: 2x^2-6x+x-3=2x(x-3)+1(x-3)=(2x+1)(x-3).
3. Cancel the common factor x-3. The answer is 2x+1.

Factorise both the top and the bottom, then look for a common bracket.

1. Factorise the numerator: two numbers that multiply to -12 and add to 1 are 4 and -3, so x^2+x-12=(x+4)(x-3).
2. The denominator is a difference of two squares: x^2-16=(x+4)(x-4).
3. Cancel the common factor x+4 to obtain \frac{x-3}{x-4}.

Compare 4-x with x-4: what do you multiply one by to get the other?

1. 4-x is the negative of x-4: 4-x=-(x-4).
2. So the fraction is \frac{-(x-4)}{x-4}. Cancel the common factor x-4.
3. The result is -1.

Length = area \div width, so write the area over the width as a fraction and simplify it.

Part (a)

1. Length = area \div width, so the length is \frac{2x^2+7x+3}{x+3}.
2. Factorise the numerator: two numbers that multiply to 2\times3=6 and add to 7 are 6 and 1, so 2x^2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3).
3. Cancel the common factor x+3: the length is (2x+1) cm.

Part (b)

1. The length is (2x+1) cm and the width is (x+3) cm (from part (a)).
2. Perimeter =2(2x+1)+2(x+3)=4x+2+2x+6=6x+8.
3. So 6x+8=38, which gives 6x=30 and x=5.
4. Check: the length is 11 cm and the width is 8 cm, and 2\times11+2\times8=38.

Distinguish a term in a sum from a factor multiplying the whole expression.

1. The numerator and denominator are sums, and x is not a factor of either whole sum.
2. Only common factors can be cancelled, so this step is invalid.
3. For example, at x=1 the original fraction is \frac{1}{2}, not \frac{3}{7}.

Factorise before substituting to keep the calculation simple.

Part (a)

1. Factorise the numerator by taking out x: x^2-4x=x(x-4).
2. The denominator is a difference of two squares: x^2-16=(x-4)(x+4).
3. Cancel the common factor x-4: R=\frac{x}{x+4}.

Part (b)

1. Use the simplified expression: R=\frac{6}{6+4}=\frac{6}{10}.
2. Simplify: R=\frac35.

Take out a common factor on the top; the bottom is a difference of two squares.

1. Factorise the numerator: a(a-b).
2. The denominator is the difference of two squares, (a-b)(a+b).
3. Cancel a-b to obtain \frac{a}{a+b}.

If the fraction simplifies to x+4, what must the numerator be?

1. If the fraction simplifies to x+4, the numerator must be (x+3)(x+4), so that x+3 cancels.
2. Expand: (x+3)(x+4)=x^2+4x+3x+12=x^2+7x+12.
3. Compare with x^2+kx+12: the x terms give k=7.

The denominator is a difference of two squares. Look for one of its factors in the numerator.

1. Numerator: two numbers that multiply to 6\times(-1)=-6 and add to 1 are 3 and -2.
2. Split the middle term: 6x^2+3x-2x-1=3x(2x+1)-1(2x+1)=(3x-1)(2x+1).
3. Denominator: 4x^2-1=(2x)^2-1^2=(2x-1)(2x+1).
4. Cancel the common factor 2x+1: \frac{3x-1}{2x-1}.

The numerator is a perfect square.

1. Numerator: 9x^2-24x+16=(3x)^2-2\times3x\times4+4^2=(3x-4)^2.
2. Denominator: two numbers that multiply to 6\times(-4)=-24 and add to -5 are -8 and 3.
3. Split the middle term: 6x^2-8x+3x-4=2x(3x-4)+1(3x-4)=(3x-4)(2x+1).
4. Cancel one factor of 3x-4: \frac{3x-4}{2x+1}.

Take out the common factor x in the numerator first.

1. Take out x in the numerator, then factorise the difference of two squares: x(x-3)(x+3).
2. The denominator is (x-3)^2.
3. Cancel one factor of x-3, leaving \frac{x(x+3)}{x-3}.

Take out the common factor 2 in the numerator first.

1. The numerator is 2(x-2)(x+2).
2. The denominator is (x-2)(x+3).
3. Cancel x-2 to obtain \frac{2(x+2)}{x+3}.

Take out a factor of -1 from the numerator first, so that its bracket starts with x^2.

1. Write the numerator with x^2 first: 6-x-x^2=-(x^2+x-6).
2. Factorise: x^2+x-6=(x+3)(x-2), so the numerator is -(x+3)(x-2)=(x+3)(2-x).
3. Denominator: two numbers that multiply to 2\times(-3)=-6 and add to 5 are 6 and -1, so 2x^2+6x-x-3=2x(x+3)-1(x+3)=(2x-1)(x+3).
4. Cancel the common factor x+3: \frac{2-x}{2x-1}, which can also be written -\frac{x-2}{2x-1}.