Question

Help me polynomials
Q3

Answer 1

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Given:

$\rm{p(x) = {2x}^{3} - 11 {x}^{2} + 17x - 6 }$

To Find:

weather 2, 3, 1/2 are zeroes of the polynomials

Solution:

• On substitution the value of p(x)=2

$\rm{\longrightarrow \: p(2) = 2( {2})^{3} - 11 {(2})^{2} + 17(2) - 6} \\ \\ \rm{\longrightarrow \:16 - 44 + 34 - 6} \\ \\ \rm{\longrightarrow 0}$

• On substitution the value of p(x)=3

$\rm \longrightarrow{p(3) = 2 {(3)}^{3} - 11 {(3)}^{2} + 17(3) - 6 }\\ \\ \rm \longrightarrow{54 - 99 + 51 - 6} \\ \\ \rm{ \longrightarrow \: 0}$

• on substitution value of p(x)= 1/2

$\rm \longrightarrow{p( \frac{1}{2} ) = 2 ({ \frac{1}{2} )}^{3} - 11 ({ \frac{1}{2}) }^{2} + 17 (\frac{1}{2}) - 6} \\ \\ \rm \longrightarrow{ \frac{1}{4} - \frac{11}{4} + \frac{17}{2} - 6} \\ \\ \rm{ \longrightarrow0}$

⚠ hence verified ⚠

hope this helps!