Exercise 9.3
Question 1. Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r (ii) ab, a – b (iii) a + b, 7a^{2}b^{2} (iv) a^{2} – 9, 4a (v) pq + qr + rp, 0
Solution:
Question 2. Complete the table.
| First expression | Second expression | Product |
^{(i)} | a | b + c + d | .... |
^{(ii)} | x + y - 5 | 5xy | .... |
(iii) | ^{p} | 6p^{2} - 7p + 5 | .... |
(iv) | _{4p}^{2}_{q}^{2} | p^{2} - q^{2} | .... |
(v) | a + b + c | abc | .... |
Solution: We have:
Question 3. Find the product.
Solution:
(iv) x*x^{2}*x^{3}*x^{4} = (1*1*1*1)*x*x^{2}*x^{3}*x^{4}
= (1)*x^{10} = x^{10}
Question 4. (a) Simplify 3x(4x – 5) + 3 and find its values for (i) x = 3 and (ii) x = 1/2 (b) Simplify a(a^{2} + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 and (iii) a = –1.
Solution:
Question 5.
(a) Add: p(p – q), q(q – r) and r(r – p)
(b) Add: 2x(z – x – y) and 2y(z – y – x)
(c) Subtract: 3l(l – 4m + 5n) from 41(10n + 3m + 2l)
(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(–a + b + c)
Solution:
(a) ∵ p(p – q) = p* p – p* q = p^{2} – pq
q(q – r) = q* q – q* r = q^{2} – qr
and r(r – p) = r* r – r* p = r2 – rp
∴ Adding the above products, we have
(p^{2} – pq) + (q^{2} – qr) + (r^{2} – rp) = p^{2} – pq + q^{2} – qr + r^{2} – rp
= p^{2} + q^{2} + r^{2} – pq – qr – rp
= p^{2} + q^{2} + r^{2} – (pq + qr + rp)
(b) ∵ 2x(z – x – y) = 2x *z – 2x *x – 2x *y = 2xz – 2x^{2} – 2xy
and 2y(z – y – x) = 2y *z – 2y *y – 2y *x = 2yz – 2y^{2} – 2yx
∴ Adding the above products, we have
(c) ∵ 3l(l – 4m + 5n) = 3l * l – 3l * 4m + 3l * 5n
= 3l^{2} – 12lm + 15ln
and 4l(10n – 3m + 2l)= 4l * 10n – 4l * 3m + 4l * 2l
= 40ln – 12lm + 8l^{2}
Multiplying a Polynomial by a Polynomial
Remember
In multiplication of polynomials with polynomials, we always look for like terms, if any, and combine them.
Exercise 9.4
Question 1. Multiply the binomials.
Solution:
Question 2. Find the product:
(i) (5 – 2x) (3 + x) (ii) (x + 7y) (7x – y)
(iii) (a^{2} + b) (a + b^{2}) (iv) (p^{2} – q^{2})(2p + q)
Solution:
(i) (5 – 2x) * (x + 3) = 5(x + 3) – 2x(x + 3)
Question 3. Simplify:
Solution:
Identities
We have studied equations earlier and we know that equations are true for certain values of the contained variables. That is the values of two sides (LHS and RHS) are equal only for certain values of variables, but it is not true for all values of the variable. If an equation is true for every value of the variable in it, then it is called an identity.
Standard Identities
Following equations are called the standard identities.
Question 1. Verify Identity (IV), for a = 2, b = 3, x = 5
Solution: We have
(x + a)(x + b) = x^{2} + (a + b)x + ab
Putting a = 2, b = 3 and x = 5 in the identity:
LHS = (x + a)(x + b)
= (5 + 2)(5 + 3)
= 7 * 8 = 56
RHS = x^{2} + (a + b)x + ab
= (5)^{2} + (2 + 3) * 5 + (2 * 3)
= 25 + (5) * 5 + 6
= 25 + 25 + 6 = 56
∴ LHS = RHS
∴ The given identity is true for the given values.
Question 2. Consider the special case of Identity (IV) with a = b, what do you get? Is it related to Identity (I)?
Solution: When a = b (each = y)
(x + a)(x + b) = x^{2} + (a + b)x + ab becomes
(x + y)(x + y) = x^{2} + (y + y)x + (y* y)
= x^{2} + (2y)x + y^{2}
= x^{2} + 2xy + y^{2}
= Yes, it is the same as Identity I.
Question 3. Consider, the special case of Identity (IV) with a = –c and b = –c. What do you get? Is it related to Identity (II)?
Solution: Identity IV is given by
(x + a)(x + b) = x^{2} + (a + b)x + ab
Replacing ‘a’ by (–c) and ‘b’ by (–c), we have
(x – c)(x – c) = x^{2} + [(–c) + (–c)] x + [(–c) * (–c)]
= x^{2} + [–2c]x + (c^{2})
= x^{2} – 2cx + c^{2}
Which is same as Identity II.
Question 4. Consider the special case of Identity (IV) with b = –a. What do you get? It is related to Identity (III).
Solution: The Identity IV is given by
(x + a)(x + b) = x^{2} + (a + b)x + ab
Replacing ‘b’ by (–a), we have:
(x + a)(x – a) = x^{2} + [a + (–a)]x + [a * (–a)]
= x^{2} + [0]x + [–a^{2}]
= x^{2 }+ 0 + (–a^{2}) = x^{2} – a^{2}
Which is same as the Identity III.