Question

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

Answer 1

C = 2

Explanation:

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