Find the rational roots of 2 x cube+ x square-7x+6
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Answers
Answer:
2x³ + x² - 7x -6 = 0. 2x³ + 2x² - x² - x - 6x - 6 = 0. 2x²(x + 1) -x(x + 1) -6(x + 1) = 0. (x + 1)(2x² - x - 6) = 0. (x + 1) ( 2x² -
16+4-14-6=0
Step-by-step explanation:
f(x) = 2x3 +x2 -7x -6
Answer:
\textbf{Given:}
f(x)=2\,x^3+x^2-7\,x-6
\textbf{To find:}
\text{Roots of f(x)}
\textbf{Solution:}
\text{Consider,}
f(x)=2\,x^3+x^2-7\,x-6
\text{Sum of the coefficients of odd powers of x}=2+(-7)=-5
\text{Sum of the coefficients of even powers of x}=1+(-6)=-5
\implies\text{Sum of the coefficients of odd powers of x}=\text{Sum of the coefficients of even powers of x}
\therefore(x+1)\;\text{is a factor of f(x)}
\textbf{By synthetic division,}
\begin{array}{r|cccc}-1&2&1&-7&-6\\&&-2&1&6\\\cline{2-5}&2&-1&-6&|\,0\end{array}
\textbf{Quotient}=2\,x^2-x-6
2\,x^2-x-6=0
2\,x^2-4x+3x-6=0
2x(x-2)+3(x-2)=0
(2x+3)(x-2)=0
\implies\,x=\dfrac{-3}{2},2
\therefore\textbf{Roots of f(x) are $\bf\,-1,\dfrac{-3}{2},2$}