Exercise 5.2: Determine the product using Formula
Required Formulae:
- $(a + b)(a – b) = a^2 – b^2$
- $(x + a)(x + b) = x^2 + (a + b)x + ab$
1. $(4x + 3), (4x – 3)$
Solution: $(4x + 3)(4x – 3)$
$= (4x)^2 – (3)^2$
$= 16x^2 – 9$
Answer:$16x^2 – 9$
2. $(13 – 12p), (13 + 12p)$
Solution: $(13 – 12p)(13 + 12p)$
$= (13)^2 – (12p)^2$
$= 169 – 144p^2$
Answer:$169 – 144p^2$
3. $(ab + 3), (ab – 3)$
Solution: $(ab + 3)(ab – 3)$
$= (ab)^2 – (3)^2$
$= a^2b^2 – 9$
Answer:$a^2b^2 – 9$
4. $(10 – xy), (10 + xy)$
Solution: $(10 – xy)(10 + xy)$
$= (10)^2 – (xy)^2$
$= 100 – x^2y^2$
Answer:$100 – x^2y^2$
5. $(4x^2 + 3y^2), (4x^2 – 3y^2)$
Solution: $(4x^2 + 3y^2)(4x^2 – 3y^2)$
$= (4x^2)^2 – (3y^2)^2$
$= 16x^4 – 9y^4$
Answer:$16x^4 – 9y^4$
6. $(a – b – c), (a + b + c)$
Solution: $(a – b – c)(a + b + c)$
$= \{a – (b + c)\}\{a + (b + c)\}$
$= a^2 – (b + c)^2$
$= a^2 – (b^2 + 2bc + c^2)$
$= a^2 – b^2 – 2bc – c^2$
Answer:$a^2 – b^2 – 2bc – c^2$
7. $(x^2 – x + 1), (x^2 + x + 1)$
Solution: $(x^2 – x + 1)(x^2 + x + 1)$
$= \{(x^2 + 1) – x\}\{(x^2 + 1) + x\}$
$= (x^2 + 1)^2 – x^2$
$= \{(x^2)^2 + 2 \cdot x^2 \cdot 1 + 1^2\} – x^2$
$= x^4 + 2x^2 + 1 – x^2$
$= x^4 + x^2 + 1$
Answer:$x^4 + x^2 + 1$
8. $\left(x – \frac{1}{2}a\right), \left(x – \frac{5}{2}a\right)$
Solution: We know,
$(x + a)(x + b) = x^2 + (a + b)x + ab$
Required product $= \left(x – \frac{1}{2}a\right)\left(x – \frac{5}{2}a\right)$
$= x^2 + \left(-\frac{1}{2}a – \frac{5}{2}a\right)x + \left(-\frac{1}{2}a\right)\left(-\frac{5}{2}a\right)$
$= x^2 + \left(\frac{-a – 5a}{2}\right)x + \frac{5}{4}a^2$
$= x^2 + \left(\frac{-6a}{2}\right)x + \frac{5}{4}a^2$
$= x^2 + (-3a)x + \frac{5}{4}a^2$
$= x^2 – 3ax + \frac{5}{4}a^2$
Answer:$x^2 – 3ax + \frac{5}{4}a^2$
9. $\left(\frac{1}{4}x – \frac{1}{3}y\right), \left(\frac{1}{4}x + \frac{1}{3}y\right)$
Solution: $\left(\frac{1}{4}x – \frac{1}{3}y\right)\left(\frac{1}{4}x + \frac{1}{3}y\right)$
$= \left(\frac{1}{4}x\right)^2 – \left(\frac{1}{3}y\right)^2$
$= \frac{1}{16}x^2 – \frac{1}{9}y^2$
Answer:$\frac{1}{16}x^2 – \frac{1}{9}y^2$
10. $(a^4 + 3a^2x^2 + 9x^4), (9x^4 – 3a^2x^2 + a^4)$
Solution: $(9x^4 + 3a^2x^2 + a^4)(9x^4 – 3a^2x^2 + a^4)$
$= \{(9x^4 + a^4) + 3a^2x^2\}\{(9x^4 + a^4) – 3a^2x^2\}$
$= (9x^4 + a^4)^2 – (3a^2x^2)^2$
$= \{(9x^4)^2 + 2 \cdot 9x^4 \cdot a^4 + (a^4)^2\} – 9a^4x^4$
$= 81x^8 + 18a^4x^4 + a^8 – 9a^4x^4$
$= 81x^8 + 9a^4x^4 + a^8$
$= a^8 + 9a^4x^4 + 81x^8$
Answer:$a^8 + 9a^4x^4 + 81x^8$
11. $(x + 1), (x – 1), (x^2 + 1)$
Solution: $(x + 1)(x – 1)(x^2 + 1)$
$= \{(x)^2 – (1)^2\}(x^2 + 1)$
$= (x^2 – 1)(x^2 + 1)$
$= (x^2)^2 – (1)^2$
$= x^4 – 1$
Answer:$x^4 – 1$
12. $(9a^2 + b^2), (3a + b), (3a – b)$
Solution: $(9a^2 + b^2)(3a + b)(3a – b)$
$= (9a^2 + b^2)\{(3a)^2 – (b)^2\}$
$= (9a^2 + b^2)(9a^2 – b^2)$
$= (9a^2)^2 – (b^2)^2$
$= 81a^4 – b^4$
Answer:$81a^4 – b^4$