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Find a number C such that the polynomial p(x)= -x+4x^2+Cx^3-8x^4

1 Answer

2

Answer:

C = 2

Step-by-step explanation:

p(x)= -x+4x^2+Cx^3-8x^4 \\  \\  \because \: given \: polynmial \: has \: zero \: at \: \\x =  \frac{1}{4}  \\  \\  \implies \: p\bigg(\frac{1}{4}  \bigg) = 0...(1) \\  \\ plug \: x = \frac{1}{4} \:  in \: p(x) \: we \: find \\  \\ p \bigg(\frac{1}{4}  \bigg)  =  - \frac{1}{4} + 4  {\bigg(\frac{1}{4} \bigg)}^{2}  + c  {\bigg(\frac{1}{4} \bigg)}^{3}  - 8  {\bigg(\frac{1}{4} \bigg)}^{4}  \\  \\ 0 = - \frac{1}{4} + \cancel 4 \times   {\frac{1}{\cancel{16} } }  + c  {\bigg(\frac{1}{64} \bigg)}  - \cancel 8  \times  \frac{1}{\cancel{256} }  \\  \\0 =\cancel{ - \frac{1}{4}} + \cancel  {{\frac{1}{4} }}  + c  {\bigg(\frac{1}{64} \bigg)}  -  \times  \frac{1}{32}  \\  \\0 =  c  {\bigg(\frac{1}{64} \bigg)}  -  \frac{1}{32}  \\  \\c  {\bigg(\frac{1}{64} \bigg)} = \frac{1}{32}   \\  \\ c = \frac{1}{32}   \times 64 \\  \\ c = 2

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Julia Schoen
15.5k 3 10 26
answered 11 months ago