If p(x)=x^{2}-4 x+3, solve : p(2)-p(-1)+pleft(frac{1}{2}right)

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2. Polynomials

medium

If $p(x)=x^{2}-4 x+3,$ evaluate $: p(2)-p(-1)+p\left(\frac{1}{2}\right)$

$\frac{31}{4}$

$\frac{-30}{5}$

$\frac{-32}{4}$

$\frac{-31}{4}$

Solution

We have $p(x)=x^{2}-4 x+3$

$\therefore p(2)-p(-1)+p\left(\frac{1}{2}\right)=\left(2^{2}-4 \times 2+3\right)-\left\{(-1)^{2}-4(-1)+3\right\}+\left\{\left(\frac{1}{2}\right)^{2}-4 \times \frac{1}{2}+3\right\}$

$=(4-8+3)-(1+4+3)+\left(\frac{1}{4}-2+3\right)$

$=-1-8+\frac{5}{4}$

$=-9+\frac{5}{4}=\frac{-36+5}{4}=\frac{-31}{4}$

Standard 9

Mathematics

From the following polynomials find out which of them has $(x-1)$ as a factor

$x^{3}+4 x^{2}+x-6$

medium

Check whether $p(x)$ is a multiple of $g(x)$ or not :

$p(x)=x^{3}-5 x^{2}+4 x-3, \quad g(x)=x-2$

easy

Expand

$\left(\frac{2 x}{3}+\frac{3 y}{4}\right)^{2}$

easy

Find the zero of the polynomial in each of the following cases

$q(y)=\pi y+3.14$

easy

Factorise :

$3 x^{3}-x^{2}-3 x+1$

hard

If $p(x)=x^{2}-4 x+3,$ evaluate $: p(2)-p(-1)+p\left(\frac{1}{2}\right)$

Solution

Similar Questions

From the following polynomials find out which of them has $(x-1)$ as a factor $x^{3}+4 x^{2}+x-6$

Check whether $p(x)$ is a multiple of $g(x)$ or not : $p(x)=x^{3}-5 x^{2}+4 x-3, \quad g(x)=x-2$

Expand $\left(\frac{2 x}{3}+\frac{3 y}{4}\right)^{2}$

Find the zero of the polynomial in each of the following cases $q(y)=\pi y+3.14$

Factorise : $3 x^{3}-x^{2}-3 x+1$

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From the following polynomials find out which of them has $(x-1)$ as a factor

$x^{3}+4 x^{2}+x-6$

Check whether $p(x)$ is a multiple of $g(x)$ or not :

$p(x)=x^{3}-5 x^{2}+4 x-3, \quad g(x)=x-2$

Expand

$\left(\frac{2 x}{3}+\frac{3 y}{4}\right)^{2}$

Find the zero of the polynomial in each of the following cases

$q(y)=\pi y+3.14$

Factorise :

$3 x^{3}-x^{2}-3 x+1$