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Question

Write the following squares of binomials as trinomials: (i) (x+2)2 (ii) (8a+3b)2 (iii) (2m+1)2 (iv) (9a+16)2 (v) (x+x22)2 (vi) (x4−y3)2 (vii) (3x−13x)2 (viii) (xy−yx)2 (ix) (3a2−5b4)2 (x) (a2b−bc2)2 (xi) (2a3b+2b3a)2 (xii) (x2−ay)2

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Solution

Using the formulas (a+b)2=a2+2ab+b2 and (a−b)2=a2−2ab+b2 (i) (x+2)2=(x)2+2×x×2+(2)2{(a+b)2=a2+2ab+b2}=x2+4x+4(ii) (8a+3b)2=(8a)2+2×8a×3b+(3b)2=64a2+48ab+9b2(iii) (2m+1)2=(2m)2+2×2m×1+(1)2=4m2+4m+1(iv) (9a+16)2=(9a)2+2×9a×16+(16)2=81a2+3a+136(v) (x+x22)2=(x)2+2×x×x22+(x22)2=x2+x3+x44(vi) (x4−y3)2=(x4)2−2×x4×y3+(y3)2=x216−16xy+y29(vii) (3x−13x)2=(3x)2−2×3x×13x+(13x)2=9x2−2+19x2(viii) (xy−yx)2=(xy)2−2×xy×yx+(yx)2=x2y2−2+y2x2(ix) (3a2−5b4)2=(3a2)2−2×32a×5b4+(5b4)2=9a24−154ab+2516b2(x) (a2b−bc2)2=(a2b)2−2a2b×bc2+(bc2)2=a4b2−2a2b2c2+b2c4(xi) (2a3b+2b3a)2=(2a3b)2+2×2a3b×2b3a+(2b3a)2=4a29b2+89+4b29a2(xii) (x2−ay)2=(x2)2−2×x2×ay+(ay)2=x4−2x2ay+a2y2

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