Question

Check whether x =
(-2/3)is one of the
zeros of polynomial
p(x) = 3x⁴ + 2x³-x²/9-x/9+2/27​

Answer 1

➲QUESTION :

• Check whether $\sf {x =\dfrac {-2}{3}}$ is one of the zeros of the polynomial $\sf{3 x^4 + 2x^3 - \dfrac {x^2}{9} - \dfrac{x}{9} + \dfrac{2}{27} .}$

☯REQUIRED KNOWLEDGE :

• If $\bf {x =\dfrac {-2}{3}}$ is one of the zeros of the given polynomial then putting this value we will get p(x) = 0 ,i.e. $\bf {p\big(\dfrac {-2}{3}\big) =0 }$.

➲SOLUTION :

Now,

$\implies \sf{p(x) = 3 x^4 + 2x^3 - \dfrac {x^2}{9} - \dfrac{x}{9} + \dfrac{2}{27}}$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = 3 \times {\bigg( \dfrac{ - 2}{3} \bigg)}^{4} + 2 \times {\bigg( \dfrac{ - 2}{3} \bigg)}^{3} - \dfrac{ {\bigg( \dfrac{ - 2}{3} \bigg)}^{2} }{9} - \dfrac{ \dfrac{ - 2}{3} }{9} + \dfrac{2}{27} }$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = 3 \times \dfrac{16}{81} + 2 \times \dfrac{ - 8}{27} - \dfrac{ \dfrac{ 4}{9} }{9} - \dfrac{ \dfrac{ - 2}{3} }{9} + \dfrac{2}{27} }$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = 3 \times \dfrac{16}{81} + 2 \times \dfrac{ - 8}{27} - \dfrac{ 4}{9} \div 9 - \bigg( \dfrac{ - 2}{3} \div 9 \bigg) + \dfrac{2}{27} }$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = 3 \times \dfrac{16}{81} + 2 \times \dfrac{ - 8}{27} - \dfrac{ 4}{9} \times \dfrac{1}{9} - \bigg( \dfrac{ - 2}{3} \times \dfrac{1}{9} \bigg) + \dfrac{2}{27} }$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = 3 \times \dfrac{16}{81} + 2 \times \dfrac{ - 8}{27} - \dfrac{ 4}{81} - \bigg (\dfrac{ -2}{27} \bigg ) + \dfrac{2}{27} }$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = \cancel{3 } \times \dfrac{16}{ \cancel{81}} + 2 \times \dfrac{ - 8}{27} - \dfrac{ 4}{81} + \dfrac{ 2}{27} + \dfrac{2}{27} }$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = \dfrac{16}{27} + \bigg(\dfrac{ - 16}{27} \bigg) - \dfrac{ 4}{81} + \dfrac{ 2}{27} \ + \dfrac{2}{27} }$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = \dfrac{16}{27} - \dfrac{ 16}{27} - \dfrac{ 4}{81} + \dfrac{ 2}{27} \ + \dfrac{2}{27} }$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = \dfrac{48 - 48 - 4 + 6 + 6}{81} }$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = \dfrac{48+ 6 + 6- 48 - 4}{81} }$

$\implies \sf{p \bigg( \dfrac{ - 2}{3} \bigg) = \dfrac{60 - 52}{81} }$

$\implies \underline{\boxed{\pink{\bf{p \bigg( \dfrac{ - 2}{3} \bigg) = \dfrac{8}{81} }}}}$

$\huge{ \pink{\therefore}}$ $\bf {x =\dfrac {-2}{3}}$ is not one of the zeros of the given polynomial .