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Mathematics

1 answer:

zmey [24]2 years ago

8 0

Answer:

2pi, 4pi, 6pi, 8pi, 10pi, 12pi, 14pi, 16pi, 18pi, 20pi

Step-by-step explanation:

Formula used=2p*r

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$x^2-1=x^2-1^2=(x-1)(x+1)$

If x²-1 is a factor of the polynomial, both x-1 and x+1 are factors of it.

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$p(1)=a \times 1^4+b \times 1^3 + c \times 1^2+ d \times 1+e \\
p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\
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a+b+c+d+e=0 \\
\underline{a-b+c-d+e=0} \\
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a+c+e=0 \\ \\
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a+b+c+d+e=0 \\
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0+b+d=0 \\
b+d=0 \\ \\
\boxed{a+c+e=b+d=0} \\
\hbox{proved } \checkmark$

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