Question 31 of 43intermediate🔧 ApplyShort Answer8 marks

Use suitable identities to find the following products:

(i)

(-3x + 4)^2

Answer

9x29x^{2} - 24x + 16

(ii)

(2s + 7)(2s - 7)

Answer

4s24s^{2} - 49

(iii)

(p2p^{2} + 1/2)(p2p^{2} - 1/2)

Answer

p4p^{4} - 1/4

(iv)

(2n + 7)(2n - 7)

Answer

4n24n^{2} - 49

(v)

(s - 2t)(s2s^{2} + 2st + 4t24t^{2})

Answer

s3s^{3} - 8t38t^{3}

(vi)

(1/2r - 4r)^2

Answer

49/4 r2r^{2}

(vii)

(-3m + 4k - l)^2

Answer

9m29m^{2} + 16k216k^{2} + l2l^{2} - 24mk - 8kl + 6lm

(viii)

(x - 1/3 y)^3

Answer

x3x^{3} - x2x^{2} y + 1/3 x y2y^{2} - 1/27 y3y^{3}

Explanation

We use standard algebraic identities to find the products without direct multiplication. For (i) and (vi), we use the identity (a - b)^2 = a2a^{2} - 2ab + b2b^{2} or (a + b)^2 = a2a^{2} + 2ab + b2b^{2}. For (ii), (iii), and (iv), we use the difference of squares identity (a + b)(a - b) = a2a^{2} - b2b^{2}. For (v), we use the identity (x - y)(x2x^{2} + xy + y2y^{2}) = x3x^{3} - y3y^{3}. For (vii), we use (a + b + c)^2 = a2a^{2} + b2b^{2} + c2c^{2} + 2ab + 2bc + 2ca. For (viii), we use (a - b)^3 = a3a^{3} - 3a2b3a^{2b} + 3ab23ab^{2} - b3b^{3}.