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+ 3 bbya+b
Each term in the first expression is multipliedbya,then
each term in the first expression is multiplied byband
the two results are added. The usual layout is shown
below.
2 a+ 3 b
a+b
Multiplying byagives 2a^2 + 3 ab
Multiplying bybgives 2 ab+ 3 b^2
Adding gives^2 a
(^2) + 5 ab+ 3 b 2
Thus,( 2 a+ 3 b)(a+b)= 2 a^2 + 5 ab+ 3 b^2
Problem 11. Multiply 3x− 2 y^2 + 4 xyby
2 x− 5 y
3 x − 2 y^2 + 4 xy
Multiplying^2 x −^5 y
by 2x→ 6 x^2 − 4 xy^2 + 8 x^2 y
Multiplying
by− 5 y→− 20 xy^2 − 15 xy+ 10 y^3
Adding gives 6 x^2 − 24 xy^2 + 8 x^2 y− 15 xy+ 10 y^3
Thus,( 3 x− 2 y^2 + 4 xy)( 2 x− 5 y)
= 6 x^2 − 24 xy^2 + 8 x^2 y− 15 xy+ 10 y^3
Problem 12. Simplify 2x÷ 8 xy
2 x÷ 8 xymeans
2 x
8 xy
2 x
8 xy
2 ×x
8 ×x×y
1 × 1
4 × 1 ×y
by cancelling
1
4 y
Problem 13. Simplify
9 a^2 bc
3 ac
9 a^2 bc
3 ac
9 ×a×a×b×c
3 ×a×c
= 3 ×a×b
= 3 ab
Problem 14. Divide 2x^2 +x−3byx− 1
(i) 2x^2 +x−3iscalledthedividendandx−1the
divisor. The usual layoutis shown below withthe
dividend and divisorboth arranged in descending
powers of the symbols.
2 x+ 3
x− 1
)
2 x^2 +x− 3
2 x^2 − 2 x
3 x− 3
3 x− 3
..
(ii) Dividing the first term of the dividend by the
first term of the divisor, i.e.
2 x^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.
2 x(x− 1 )= 2 x^2 − 2 x, which is placedunder 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 )= 3 x−3, which is placedunder
the3x−3. The remainder, on subtraction, is zero,
which completes the process.
Thus,( 2 x^2 +x− 3 )÷(x− 1 )=( 2 x+ 3 ).
(A check can be made on this answer by
multiplying( 2 x+ 3 )by(x− 1 ), which equals
2 x^2 +x−3.)
Problem 15. Simplify
x^3 +y^3
x+y