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Question

Find the value of the expression 2x3+2x27x+72 where x=3+512.

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Solution

It is given that x=3+512 which is same as x=3+5i2 because i2=1.

Now, let us substitute x=3+5i2 in the given polynomial 2x3+2x27x+72 as follows:

2x3+2x27x+72=2(3+5i2)3+2(3+5i2)27(3+5i2)+72=2(33+(5i)3+(3×32×5i)+(3×3×(5i)2)8)+2(32+(5i)2+(2×3×5i)2)(21+35i2)+72((a+b)3=a3+b3+3a2b+3ab2,(a+b)2=a2+b2+2ab)=2(27125i+135i2258)+2(925+30i4)(21+35i2)+72(i2=1)
=(198+10i4)+(16+30i2)(21+35i2)+72=(2(99+5i)4)+(30i162)(35i+212)+72=5i992+30i162(35i+212)+(72×22)=5i99+30i1635i21+1442=35i35i136+1442=82=4

Hence, 2x3+2x27x+72=4 when x=3+512.

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