Test Index

BECE Mathematics 2021 Paper

© africaexams.com

Question : 28

Total: 40

Expand

(7 r-5)(3 r+4)

$21 r^2+13 r-20$
$21 r^2-13 r-20$
$21 r^2-43 r-20$
$21 r^2+43 r-20$

Solution:

Expand simply means multiply and simplify

Principle

\;(a+b)(c+d)=a(c+d)+b(c+d)

\;(a-b)(c+d)=a(c+d)-b(c+d)

\;(7 r-5)(3 r+4)=7 r(3 r+4)-5(3 r+4)

\;7 r(3 r+4)-5(3 r+4)=7 r \times 3 r+7 r \times 4-5 \times 3 r-5 \times 4

\;7 r \times 3 r+7 r \times 4-5 \times 3 r-5 \times 4=21 r^2+28 r-15 r-20

\;21 r^2+28 r-15 r-20=21 r^2+13 r-20

NOTE:

r \times r=r^2

a=a^1

From the law of indices,

a^m \times a^n=a^{m+n}

r \times r=r^1 \times r^1=r^{1+1}=r^2

© africaexams.com

Go to Question:

Prev Question

Next Question