Exercise 11.3

1. Multiply the binomials:

(i) 2a-9 and 3a+4

Solution:

(2a - 9)(3a + 4) = 2a(3a + 4) - 9(3a + 4)

= 6a26a + a - 27a -

= 6a2 - a - 36

(ii) x-2y and 2x-y

Solution:

(x - 2y)(2x - y)

= x(2x - y) - 2y(2x - y)

= 2x22x - - + 2y22y3

= 2x2 - + 2y2

(iii) kl+lm and k-l

Solution:

(kl + lm)(k - l)

= kl(k - l) + lm(k - l)

= k2lk3l - kl2k2l2 + klmkl - lm2l2m2

(iv) m2−n2 and m+n

Solution:

(m2−n2)(m + n)

= m2(m + n) - n2(m + n)

= m33m3 + m2nm2n2 - mn2mn3 - n33n3

2. Find the product:

(i) (x+y)(2x-5y+3xy)

Solution:

(x + y)(2x - 5y + 3xy)

= x(2x - 5y + 3xy) + y(2x - 5y + 3xy)

= 2x22x - 5xy + 3xy + 2xy2y - 5y2 +

= 2x2 - 3xy + 3x2y3x2y2 - 5y2 + 3xy2

(ii) a−2b+3cab2−a2b

Solution:

a−2b+3cab2−a2b

= aab2−a2b - 2bab2−a2b + 3cab2−a2b

= a2b22a2b2 - a3ba2b - 2ab32a3b3 + 2a2b22a3b2 + 3ab3c3ab3c3 - 3a3bc3ab3c

= 3a2b2a2b2 - a3b2a3b - 2ab33ab3 + 3ab3c3ab3c3 - 3a3bc4a3b3c

(iii) (mn-kl+km)(kl-lm)

Solution:

(mn - kl + km)(kl - lm)

= mn(kl - lm) - kl(kl - lm) + km(kl - lm)

= mnkl - m2ln - k2l2k2l + kl2mkl3m + k2lmk2l2m - klm2kl2m2

= mnkl - m2ln - k2l2 + kl2m + k2lm - klm2

(iv) p3+q3p−5q+6r

Solution:

p3+q3p−5q+6r

= p3p−5q+6r + q3p−5q+6r

= p4p5 - 5p3q5p2q2 + 6p3r6p4r2 + pq3p2q2 - 5q45q3 + 6q3r26q2r2

= p4 - 5p3q + 6p3r + pq3 - 5q4 + 6q3r

3. Simplify the following : (x-2y)(y-3x)+(x+y)(x-3y)-(y-3x)(4x-5y)

Solution:

(x - 2y)(y - 3x) + (x + y)(x - 3y) - (y - 3x)(4x - 5y) = (xyx2y - 3x23x - 2y22y3 + 6xy6x) + (x2x3 - 3xy3x2y + xy2x2y - 3y23y3) - (4xy2xy - 5y25y3 - 12x212x + 15xy15x2y)

= (7xy - 3x2 - 2y2) + (x2 - 2xy - 3y2) - (19xy - 12x2 - 5y2) = 7xy - 3x2 - 2y2 + x2 - 2xy - 3y2 - 19xy + 12x2 + 5y2

= −3x2+x2+12x2 + 7xy−2xy−19xy + −2y2−3y2+5y2

= x2 - xy + y2

(ii) m+nm2−mn+n2

Solution:

m+nm2−mn+n2

= mm2−mn+n2+nm2−mn+n2

= m3m4 - m2nm2 + mn2mn3 + m2nm3n - mn2m2n3 + n3n2

= m3m2n + n3mn2

(iii) (a-2b+5c)(a-b) - (a-b-c)(2a+3c) + (6a+b)(2c-3a-5b)

Solution:

(a - 2b + 5c)(a - b) - (a - b - c)(2a + 3c) + (6a + b)(2c - 3a - 5b)

= (a2a3 - ab2a - 2ab2b + 2b22b3 + 5ac5a - 5bc5c) - (2a2a2 + 3ac3c - 2ab2b - 3bc3b - 2ac2c - 3c23c3|) + (12ac6ac - 18a218a3 - 30ab115ab + 2bcbc - 3ab5ab - 5b25b3)

= a2−3ab+2b2+5ac−5bc - 2a2+ac−2ab−3bc−3c2 + 12ac−18a2−33ab+2bc−5b2

= a2−3ab+2b2+5ac−5bc−2a2−ac+2ab+3bc+3c2+12ac−18a2−33ab+2bc−5b2

= a2−2a2−18a2 + 2b2−5b2 + 3c2 + −3ab+2ab−33ab + 5ac−ac+12ac + −5bc+3bc+2bc

= a2 + b2 + c2 - ab + ac

(iv) (pq-qr+pr)(pq+qr) - (pr+pq)(p+q-r)

Solution:

(pq - qr + pr)(pq + qr) - (pr + pq)(p + q - r)

= (p2q2pq2 - q2rq2r + p2r2p2r2) - (p2rp3r2 + pqrpq - pr2p2r2 + p2qp3q + pq2p2q2 - pqrpq2r)

= p2q2 - q2r2 + p2r2 - p2r - pqr + pr2 - p2q - pq2 + pqr

= p2q22p2q2 + −q2r2q2r2 + 2p2r2p2r2 + −p2r−2p2r + −p2qp2q + −pq2pq2