Worked examples · Sheet 1

Special products and factoring out common factors

Theory used: Square and cube of a binomial Difference of squares Sum and difference of cubes Factoring out a common factor and factoring by grouping

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Expand $(3 x - 2)^{2}$ and $(x + 2)^{3}$ .

(3 x - 2)^{2} = (3 x)^{2} - 2 \cdot 3 x \cdot 2 + 2^{2} = 9 x^{2} - 12 x + 4

(x + 2)^{3} = x^{3} + 3 \cdot x^{2} \cdot 2 + 3 \cdot x \cdot 2^{2} + 2^{3} = x^{3} + 6 x^{2} + 12 x + 8

In the square of the binomial the bases are $A = 3 x$ and $B = 2$ , so the squares are $9 x^{2}$ and $4$ and the double product is $2 \cdot 3 x \cdot 2 = 12 x$ ; the binomial contains a minus sign, therefore the double product appears with a negative sign.

In the cube of the binomial the bases are $A = x$ and $B = 2$ . The four terms of the expansion are $A^{3} = x^{3}$ , $3 A^{2} B = 3 \cdot x^{2} \cdot 2 = 6 x^{2}$ , $3 A B^{2} = 3 \cdot x \cdot 4 = 12 x$ and $B^{3} = 8$ ; the binomial is a sum, so all the signs are positive.

Exercise 2

Factor $4 x^{2} - 25$ and $8 x^{3} - 27$ .

4 x^{2} - 25 = (2 x)^{2} - 5^{2} = (2 x - 5) (2 x + 5)

8 x^{3} - 27 = (2 x)^{3} - 3^{3} = (2 x - 3) (4 x^{2} + 6 x + 9)

The binomial $4 x^{2} - 25$ is the difference of the squares of $2 x$ and $5$ , so it factors as the product of the difference and the sum of the bases.

The binomial $8 x^{3} - 27$ is the difference of the cubes of $A = 2 x$ and $B = 3$ , therefore the formula $A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})$ applies: the first factor is $2 x - 3$ , the second is $A^{2} + A B + B^{2} = 4 x^{2} + 6 x + 9$ . The second factor is a false square with no real roots, so the factorization is complete.

Exercise 3

Factor $2 x^{3} - 8 x$ and $3 x^{3} y + 6 x^{2} y^{2} + 3 x y^{3}$ .

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

3 x^{3} y + 6 x^{2} y^{2} + 3 x y^{3} = 3 x y (x^{2} + 2 x y + y^{2}) = 3 x y (x + y)^{2}

In the first polynomial all the terms contain the factor $2 x$ , so the first step is factoring out the common factor $2 x$ . The remaining factor $x^{2} - 4$ is a difference of squares, therefore it factors as $(x - 2) (x + 2)$ and the factorization is complete.

In the second polynomial the common factor is $3 x y$ , and factoring it out leaves $x^{2} + 2 x y + y^{2}$ . This trinomial contains the squares of $x$ and $y$ and the double product $2 x y$ with a positive sign, so it is the square of the binomial $x + y$ .