Basic algebra 63
= 5 × 2 ×
2
5
2
×
5
2
3
since 2
1
2
=
5
2
=
5
1
×
2
1
×
2
5
×
2
5
×
5
2
×
5
2
×
5
2
=
1
1
×
1
1
×
1
1
×
1
1
×
1
1
×
5
1
×
5
1
by cancelling
= 5 × 5
= 25
Problem 10. Multiply 2a + 3b by a + b
Each term in the first expression is multiplied by a, then
each term in the first expression is multiplied by b and
the two results are added. The usual layout is shown
below.
2a + 3b
a + b
Multiplying by a gives
2a 2 + 3ab
Multiplying by b gives
2ab + 3b 2
Adding gives
2a 2 + 5ab + 3b 2
Thus, (2a + 3b)(a + b) = 2a
2
+ 5ab + 3b
2
Problem 11. Multiply 3x − 2y 2 + 4x y by
2x − 5y
3x − 2y 2 + 4x y
2x − 5y
Multiplying
by 2x →
6x 2 − 4x y 2 + 8x 2 y
Multiplying
by −5y →
−20x y 2
−15x y + 10y 3
Adding gives 6x 2 − 24xy 2 + 8x 2 y − 15xy + 10y 3
Thus, (3x − 2y 2 + 4xy)(2x − 5y)
= 6x 2 − 24xy 2 + 8x 2 y − 15xy + 10y 3
Problem 12. Simplify 2x ÷ 8x y
2x ÷ 8x y means
2x
8x y
2x
8x y
=
2 × x
8 × x × y
=
1 × 1
4 × 1 × y
by cancelling
=
1
4y
Problem 13. Simplify
9a
2 bc
3ac
9a 2 bc
3ac
=
9 × a × a × b × c
3 × a × c
= 3 × a × b
= 3ab
Problem 14. Divide 2x 2 + x − 3 by x − 1
(i) 2x 2 + x − 3 is called the dividend and x − 1 the
divisor. The usual layout is shown below with the
dividend and divisor both arranged in descending
powers of the symbols.
2x + 3
x − 1
2x 2 + x − 3
2x 2 − 2x
3x − 3
3x − 3
.
.
(ii) Dividing the first term of the dividend by the
first term of the divisor, i.e.
2x 2
x
gives 2x, which
is put above the first term of the dividend as
shown.
(iii) The divisor is then multiplied by 2x, i.e.
2x(x − 1) = 2x 2 − 2x, which is placed under the
dividend as shown. Subtracting gives 3x − 3.
(iv) The process is then repeated, i.e. the first term
of the divisor, x, is divided into 3x, giving +3,
which is placed above the dividend as shown.
(v) Then 3(x − 1) = 3x − 3, which is placed under
the 3x − 3. The remainder, on subtraction, is zero,
which completes the process.
Thus, (2x 2 + x − 3) ÷ (x − 1)=(2x + 3).
(A check can be made on this answer by
multiplying (2x + 3) by (x − 1), which equals
2x 2 + x − 3.)
Problem 15. Simplify
x
3
+ y
3
x + y
= 5 × 2 ×
2
5
2
×
5
2
3
since 2
1
2
=
5
2
=
5
1
×
2
1
×
2
5
×
2
5
×
5
2
×
5
2
×
5
2
=
1
1
×
1
1
×
1
1
×
1
1
×
1
1
×
5
1
×
5
1
by cancelling
= 5 × 5
= 25
Problem 10. Multiply 2a + 3b by a + b
Each term in the first expression is multiplied by a, then
each term in the first expression is multiplied by b and
the two results are added. The usual layout is shown
below.
2a + 3b
a + b
Multiplying by a gives
2a 2 + 3ab
Multiplying by b gives
2ab + 3b 2
Adding gives
2a 2 + 5ab + 3b 2
Thus, (2a + 3b)(a + b) = 2a
2
+ 5ab + 3b
2
Problem 11. Multiply 3x − 2y 2 + 4x y by
2x − 5y
3x − 2y 2 + 4x y
2x − 5y
Multiplying
by 2x →
6x 2 − 4x y 2 + 8x 2 y
Multiplying
by −5y →
−20x y 2
−15x y + 10y 3
Adding gives 6x 2 − 24xy 2 + 8x 2 y − 15xy + 10y 3
Thus, (3x − 2y 2 + 4xy)(2x − 5y)
= 6x 2 − 24xy 2 + 8x 2 y − 15xy + 10y 3
Problem 12. Simplify 2x ÷ 8x y
2x ÷ 8x y means
2x
8x y
2x
8x y
=
2 × x
8 × x × y
=
1 × 1
4 × 1 × y
by cancelling
=
1
4y
Problem 13. Simplify
9a
2 bc
3ac
9a 2 bc
3ac
=
9 × a × a × b × c
3 × a × c
= 3 × a × b
= 3ab
Problem 14. Divide 2x 2 + x − 3 by x − 1
(i) 2x 2 + x − 3 is called the dividend and x − 1 the
divisor. The usual layout is shown below with the
dividend and divisor both arranged in descending
powers of the symbols.
2x + 3
x − 1
2x 2 + x − 3
2x 2 − 2x
3x − 3
3x − 3
.
.
(ii) Dividing the first term of the dividend by the
first term of the divisor, i.e.
2x 2
x
gives 2x, which
is put above the first term of the dividend as
shown.
(iii) The divisor is then multiplied by 2x, i.e.
2x(x − 1) = 2x 2 − 2x, which is placed under the
dividend as shown. Subtracting gives 3x − 3.
(iv) The process is then repeated, i.e. the first term
of the divisor, x, is divided into 3x, giving +3,
which is placed above the dividend as shown.
(v) Then 3(x − 1) = 3x − 3, which is placed under
the 3x − 3. The remainder, on subtraction, is zero,
which completes the process.
Thus, (2x 2 + x − 3) ÷ (x − 1)=(2x + 3).
(A check can be made on this answer by
multiplying (2x + 3) by (x − 1), which equals
2x 2 + x − 3.)
Problem 15. Simplify
x
3
+ y
3
x + y