Question

check weather 1/2, 1 , -1/2 are the zeroes of the polynomial 2x^2 + x^2- 5x + 2​

Answer 1

$\text{Given polynomial p(x) = } 2x^3 + x^2 - 5x + 2\\\\\text{Zeros to check = } \dfrac{1}{2}, \: 1, \: \dfrac{-1}{2}\\\\\text{Let's begin.}\\\\p \left( \dfrac{1}{2} \right) = 2 \left( \dfrac{1}{2} \right) ^3 + \left( \dfrac{1}{2} \right) ^2 - 5\left( \dfrac{1}{2} \right) + 2\\\\\\= 2 \left( \dfrac{1}{8} \right) + \left( \dfrac{1}{4} \right) - \left( \dfrac{5}{2} \right) + 2\\\\\\= \left( \dfrac{1}{4} \right) + \left( \dfrac{1}{4} \right) - \left( \dfrac{5}{2} \right) + 2$

$\left( \dfrac{1+1-10+8}{4} \right)\\\\\\= \left( \dfrac{10-10}{4} \right)\\\\\\= \left( \dfrac{0}{4} \right)\\\\\\= 0\\\\\text{Thus, \dfrac{1}{2} is a zero of p(x).}\\\\\text{Now, taking p(1), we have}\\\\2(1)^3 + 1^2 - 5(1) + 2\\\\= 2 + 1 - 5 + 2 \\\\= 5-5\\\\= 0\\\\\text{Thus, 1 is also a zero of p(x).}\\\\\text{Let's ckeck - \dfrac{1}{2} now.}$

$p \left( \dfrac{-1}{2} \right) = 2 \left( \dfrac{-1}{2} \right) ^3 + \left( \dfrac{-1}{2} \right) ^2 - 5 \left( \dfrac{-1}{2} \right) + 2\\\\\\= 2 \left( \dfrac{-1}{8} \right) + \left( \dfrac{1}{4} \right) + \left( \dfrac{5}{2} \right) + 2\\\\\\= \left( \dfrac{-1}{4} \right) + \left( \dfrac{1}{4} \right) + \left( \dfrac{5}{2} \right) + 2\\\\\\= \left( \dfrac{5}{2} \right) + 2\\\\\\= \left( \dfrac{5+4}{2} \right)\\\\\\= \left( \dfrac{9}{2} \right) \ne 0$

$\text{Thus, \dfrac{-1}{2} is not a zero of p(x).}$

Answer 2

$\huge{\boxed{\textbf{Answer}}}$

♦ Well in the question above it is asked to check whether $\dfrac{1}{2}$ , 1 and $\dfrac{-1}{2}$ are the zero of the polynomial $2x^2 + x^2 - 5x + 2$

Well the Polynomial as the terms $2x^2 \: and\: x^2$ is away we will consider it as = $2x^3 + x^2 - 5x + 2$

♦ Now before we find out we must know what is the zero of a Polynomial ?

>> Well the zero of a Polynomial is the value of "x" for which the end product of the polynomial is zero .

♦ For checking we will substitute the value of x as $\dfrac{1}{2}$ , 1 and $\dfrac{-1}{2}$

$\large{\boxed{\textbf{Checking}}}$

♦ When x = $\dfrac{1}{2}$

• Then

$2x^2 +x^2 - 5x + 2$

$= 2(\dfrac{1}{2})^3 + (\dfrac{1}{2})^2 - 5(\dfrac{1}{2})+ 2$

$= \dfrac{2}{8} + \dfrac{1}{4} - \dfrac{5}{2}+ 2$

$= \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{5}{2} +2$

$= \dfrac{2}{4} - \dfrac{5}{2} + 2$

$= \dfrac{1}{2} - \dfrac{5}{2} + 2$

$= \dfrac{1-5+4}{2}$

$= \dfrac{5 - 5}{2}$

$= \dfrac{0}{2}$

= 0

:- Hence a zero .

♦ When x = 1

• Then

$2x^2 +x^2 -5x +2$

$= 2(1^3) + (1^2) - 5(1) + 2$

$= 2 + 1 - 5 + 2$

$= 3 + 2 - 5$

$= 5 - 5$

:- Hence a zero .

♦ When x = $\dfrac{-1}{2}$

• Then

$2x^3 + x^2 - 5x + 2$

$= 2(\dfrac{-1}{2})^3 + (\dfrac{1}{2})^2 - 5(\dfrac{-1}{2}) + 2$

$= \dfrac{-2}{8} + \dfrac{1}{4} - \dfrac{-5}{2} + 2$

$= \dfrac{-1}{4} + \dfrac{1}{4} - \dfrac{-5}{2} + 2$

$= \dfrac{5}{2} +2$

$= \dfrac{5+4}{2}$

$= \dfrac{9}{2}$

:- Hence not a zero .