Find k such that the polynomial $$P(x)=-2{x}^{3}+k{x}^{2}+2x+4$$ is divisible by x-2.

Keyla Koch
2022-10-24
Answered

Find k such that the polynomial $$P(x)=-2{x}^{3}+k{x}^{2}+2x+4$$ is divisible by x-2.

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Dobricap

Answered 2022-10-25
Author has **14** answers

$$P(x)=-2{x}^{3}+k{x}^{2}+2x+4$$

As $$(x-1)$$ is a factor of polynomial $$p(x)=-2{x}^{3}+kx+2x+4$$

So $$p(2)=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2(2{)}^{3}+k(2{)}^{2}+2(2)+4=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2(8)+k(4)+4+4=0\phantom{\rule{0ex}{0ex}}\Rightarrow 16+4k+8=0\phantom{\rule{0ex}{0ex}}\Rightarrow -8+4k=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4k=8\phantom{\rule{0ex}{0ex}}k=\frac{8}{4}\phantom{\rule{0ex}{0ex}}k=2$$

As $$(x-1)$$ is a factor of polynomial $$p(x)=-2{x}^{3}+kx+2x+4$$

So $$p(2)=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2(2{)}^{3}+k(2{)}^{2}+2(2)+4=0\phantom{\rule{0ex}{0ex}}\Rightarrow -2(8)+k(4)+4+4=0\phantom{\rule{0ex}{0ex}}\Rightarrow 16+4k+8=0\phantom{\rule{0ex}{0ex}}\Rightarrow -8+4k=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4k=8\phantom{\rule{0ex}{0ex}}k=\frac{8}{4}\phantom{\rule{0ex}{0ex}}k=2$$

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a) Find quadratic and cubic Taylor polynomials for $x}^{\frac{1}{3}$ at x=1

b) Show that these polynomials are upper and lower bounds for$x}^{\frac{1}{3}$ for x>1

c) Use these bounds to approximate$1.2}^{\frac{1}{3}$ and $2\frac{1}{3}$

b) Show that these polynomials are upper and lower bounds for

c) Use these bounds to approximate