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Question 15 Marks
Evaluate:
(i) $(a+b)(a-b)$
(ii) $\left(a^2+b^2\right)(a+b)(a-b)$; using the result of (i).
(iii) $\left(a^4+b^4\right)\left(a^2+b^2\right)(a+b)(a-b)$; using the result of (ii).
Answer
$(i) (a + b)(a - b)$
$= a (a - b) + b(a - b)$
$= a^2 - ab + ab - b^2$
$= a^2 - b^2$
$(ii) (a^2 + b^2)(a + b)(a - b)$
$= (a^2 + b^2)(a^2 - b^2) ...{from(i)}$
$= a2 (a^2 - b^2) + b^2 (a^2 - b^2)$
$= a^4 - a^2b^2 + a^2b^2 - b^4$
$= a^8 - a^4b^4 + a^4b^4 - b^8$
$= a^4 - b^4$
$(iii) (a^4 + b^4)(a^2 + b^2)(a + b)(a - b)$
$= (a^4 + b^4) (a^4 - b^4) ....{from(ii)}$
$= a^4 (a^4 + b^4) + b^4 (a^4 + b^4)$
$= a^8 - a^4b^4 + a^4b^4 - b^8$
$= a^8 - b^8$
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Question 25 Marks
Subtract the sum of $3 a^2-2 a+5$ and $a^2-5 a-7$ from the sum of $5 a^2-9 a+3$ and $2 a-a^2-1$
Answer
Sum of $3 a^2-2 a+5$ and $a^2-5 a-7$
$= 3a^2 – 2a + 5 + a^2 – 5a – 7$
$= 3a^2 + a^2 – 2a – 5a + 5 – 7$
$= 4a^2 - 7a - 2$
and sum of $5a^2 -9a + 3$ and $2a – a^2 – 1$
$= 5a^2 - 9a + 3 + 2a – a^2 – 1$
$= 5a^2 – a^2 - 9a + 2a + 3 - 1$
$= 4a^2 - 7a + 2$
Now $(4a^2 - 7a + 2) - (4a^2 - 7a - 2)$
$= 4a^2 - 7a + 2 - 4a^2 + 7a + 2$
$= 4a^2 - 4a^2 - 7a + 7a + 2 + 2$
$= 0 + 0 + 4 = 4$
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