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Question

Mathematics Question on Binomial theorem

Expand (1−x+x2)4(1 - x + x^2)^4.

A

1−4x+10x2−16x3+19x4−16x2+10x6−4x7+x8 1-4 x + 10x^{2}- 16x^{3}+ 19x^{4}- 16x^{2}+ 10x^{6}-4x^{7} + x^{8}

B

1−4x+10x2−16x3+x4+16x5+10x6+4x7−x81-4 x + 10x^{2}- 16x^{3}+ x^{4}+ 16x^{5}+ 10x^{6}+4x^{7} - x^{8}

C

1−4x−10x2+16x3−19x4+16x5−10x6+4x7+x81-4 x -10x^{2}+ 16x^{3}-19 x^{4}+ 16x^{5}- 10x^{6}+4x^{7} + x^{8}

D

1−4x−10x2−16x3+19x4−16x5+x81-4 x -10x^{2}- 16x^{3}+ 19x^{4}- 16x^{5} + x^{8}

Answer

1−4x+10x2−16x3+19x4−16x2+10x6−4x7+x8 1-4 x + 10x^{2}- 16x^{3}+ 19x^{4}- 16x^{2}+ 10x^{6}-4x^{7} + x^{8}

Explanation

Solution

Put 1−x=y1 - x = y, then (1−x+x2)4=(y+x2)44(1 - x + x^2)^4 = (y + x^2)^44 = 4C0 y4+ 4C1 y3(x2)1+ 4C2 y2(x2)2+ 4C3 y(x2)3=\,^{4}C_{0}\,y^{4}+\,^{4}C_{1}\,y^{3}\left(x^{2}\right)^{1}+\,^{4}C_{2}\,y^{2}\left(x^{2}\right)^{2}+\,^{4}C_{3}\,y\left(x^{2}\right)^{3} + 4C4(x2)4+\,^{4}C_{4}\left(x^{2}\right)^{4} =y4+4y3 x2+6y2 x4+4y x6+x8= y^{4} + 4y^{3}\, x^{2} + 6y^{2}\, x^{4} + 4y\, x^{6} + x^{8} =(1−x)4+4x2(1−x)3+6x4(1−x)2+4x6(1−x)+x8= \left(1 - x\right)^{4} + 4x^{2}\left(1 - x\right)^{3} + 6x^{4}\left(1 - x\right)^{2} + 4x^{6}\left(1 - x\right) + x^{8} =1−4x+10x2−16x3+19x4−16x2+10x6−4x7+x8= 1-4 x + 10x^{2}- 16x^{3}+ 19x^{4}- 16x^{2}+ 10x^{6}-4x^{7} + x^{8}