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Marianna [84]
3 years ago
8

Find the coefficient of the 4th term in expansion of (x +2 )^5

Mathematics
1 answer:
LuckyWell [14K]3 years ago
6 0
(x + 2)^5=(x+2)^2\cdot{(x+2)^2}\cdot{(x+2)}\\\\(x+2)^5=(x^2+4x+4)\cdot{(x^2+4x+4)}\cdot(x+2)

(x+2)^5=(x^4+8x^3+24x^2+32x+16)\cdot(x+2)

(x+2)^5=x^5+10x^4+40x^3+80x^2+80x+32

10

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6x ^2-7x+2=0\\ \\ a=6 , \ b = -7 , \ c=2 \\ \\\Delta = b^{2}-4ac = (-7)^{2}-4*6*2=49-48=1 \\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{7-\sqrt{1}}{2*6}=\frac{7-1}{12} =\frac{6}{12}= \frac{1}{ 2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{7+\sqrt{1}}{2*6}=\frac{7+1}{12} =\frac{8}{12}= \frac{2}{ 3}


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