Question

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

Answer 1

Answer:

C = 2

Step-by-step explanation:

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