
Given that \[x - 3\] and \[2x + 1\] are factors of \[2{x^3} + p{x^2} + \left( {2q - 1} \right)x + q\] :
Find the value of p and the value of q.
Answer
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Hint: In this question since the two factors of the quadratic equation are given so we will substitute these roots one by one in the given polynomial equation and then we will solve the equation for the value of p and q from the equation we obtained by substituting the values of the factors in the given equation.
Complete step-by-step answer:
Given the factors of polynomial
\[x - 3\]
\[2x + 1\]
Let the function be
\[f(x) = 2{x^3} + p{x^2} + \left( {2q - 1} \right)x + q\]
We can write the roots as
\[
x - 3 = 0 \\
x = 3 \\
\]
\[
2x + 1 = 0 \\
x = - \dfrac{1}{2} \\
\]
Now we substitute the value of the root \[x = 3\] in the function\[f(x)\] , so we get
\[
\Rightarrow f(3) = 2{\left( 3 \right)^3} + p{\left( 3 \right)^2} + \left( {2q - 1} \right) \times 3 + q \\
= 2 \times 27 + p \times 9 + 3\left( {2q - 1} \right) + q \\
= 54 + 9p + 6q - 3 + q \\
= 10p + 6q + 51 - - (i) \;
\]
When\[x = - \dfrac{1}{2}\] , we get the equation
\[
\Rightarrow f\left( { - \dfrac{1}{2}} \right) = 2{\left( { - \dfrac{1}{2}} \right)^3} + p{\left( { - \dfrac{1}{2}} \right)^2} + \left( {2q - 1} \right)\left( { - \dfrac{1}{2}} \right) + q \\
= 2\left( { - \dfrac{1}{8}} \right) + \dfrac{1}{4}p - q + \dfrac{1}{2} + q \\
= - \dfrac{1}{4} + \dfrac{1}{4}p + \dfrac{1}{2} \\
= \dfrac{1}{4} + \dfrac{1}{4}p - - (ii) \;
\]
Now by solving equation (ii),
\[
\Rightarrow f\left( { - \dfrac{1}{2}} \right) = 0 \\
\Rightarrow \dfrac{1}{4} + \dfrac{1}{4}p = 0 \\
\Rightarrow p = - 1 \\
\]
We get the value of\[p = - 1\] ,
Now we substitute the value of p in the equation (i),
\[
\Rightarrow 10p + 6q + 51 = 0 \\
\Rightarrow 10\left( { - 1} \right) + 6q + 51 = 0 \\
\Rightarrow - 10 + 6q + 51 = 0 \\
\Rightarrow 6q = - 41 \\
\Rightarrow q = \dfrac{{ - 41}}{6} \;
\]
We get the value of\[q = \dfrac{{ - 41}}{6}\] ,
Hence the value of p and the value of q is
\[p = - 1\]
\[q = \dfrac{{ - 41}}{6}\]
Note: Polynomial is an expression consisting of variables and coefficients which involves the operation of addition, subtraction, multiplication and non-negative integer exponentiation of variables.
Complete step-by-step answer:
Given the factors of polynomial
\[x - 3\]
\[2x + 1\]
Let the function be
\[f(x) = 2{x^3} + p{x^2} + \left( {2q - 1} \right)x + q\]
We can write the roots as
\[
x - 3 = 0 \\
x = 3 \\
\]
\[
2x + 1 = 0 \\
x = - \dfrac{1}{2} \\
\]
Now we substitute the value of the root \[x = 3\] in the function\[f(x)\] , so we get
\[
\Rightarrow f(3) = 2{\left( 3 \right)^3} + p{\left( 3 \right)^2} + \left( {2q - 1} \right) \times 3 + q \\
= 2 \times 27 + p \times 9 + 3\left( {2q - 1} \right) + q \\
= 54 + 9p + 6q - 3 + q \\
= 10p + 6q + 51 - - (i) \;
\]
When\[x = - \dfrac{1}{2}\] , we get the equation
\[
\Rightarrow f\left( { - \dfrac{1}{2}} \right) = 2{\left( { - \dfrac{1}{2}} \right)^3} + p{\left( { - \dfrac{1}{2}} \right)^2} + \left( {2q - 1} \right)\left( { - \dfrac{1}{2}} \right) + q \\
= 2\left( { - \dfrac{1}{8}} \right) + \dfrac{1}{4}p - q + \dfrac{1}{2} + q \\
= - \dfrac{1}{4} + \dfrac{1}{4}p + \dfrac{1}{2} \\
= \dfrac{1}{4} + \dfrac{1}{4}p - - (ii) \;
\]
Now by solving equation (ii),
\[
\Rightarrow f\left( { - \dfrac{1}{2}} \right) = 0 \\
\Rightarrow \dfrac{1}{4} + \dfrac{1}{4}p = 0 \\
\Rightarrow p = - 1 \\
\]
We get the value of\[p = - 1\] ,
Now we substitute the value of p in the equation (i),
\[
\Rightarrow 10p + 6q + 51 = 0 \\
\Rightarrow 10\left( { - 1} \right) + 6q + 51 = 0 \\
\Rightarrow - 10 + 6q + 51 = 0 \\
\Rightarrow 6q = - 41 \\
\Rightarrow q = \dfrac{{ - 41}}{6} \;
\]
We get the value of\[q = \dfrac{{ - 41}}{6}\] ,
Hence the value of p and the value of q is
\[p = - 1\]
\[q = \dfrac{{ - 41}}{6}\]
Note: Polynomial is an expression consisting of variables and coefficients which involves the operation of addition, subtraction, multiplication and non-negative integer exponentiation of variables.
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