Question

find rational root of the polynomial f(x)=2x3+x2-7x-6

Answer 1

$\textbf{Given:}$

$f(x)=2\,x^3+x^2-7\,x-6$

$\textbf{To find:}$

$\text{Roots of f(x)}$

$\textbf{Solution:}$

$\text{Consider,}$

$f(x)=2\,x^3+x^2-7\,x-6$

$\text{Sum of the coefficients of odd powers of x}=2+(-7)=-5$

$\text{Sum of the coefficients of even powers of x}=1+(-6)=-5$

$\implies\text{Sum of the coefficients of odd powers of x}=\text{Sum of the coefficients of even powers of x}$

$\therefore(x+1)\;\text{is a factor of f(x)}$

$\textbf{By synthetic division,}$

$\begin{array}{r|cccc}-1&2&1&-7&-6\\&&-2&1&6\\\cline{2-5}&2&-1&-6&|\,0\end{array}$

$\textbf{Quotient}=2\,x^2-x-6$

$2\,x^2-x-6=0$

$2\,x^2-4x+3x-6=0$

$2x(x-2)+3(x-2)=0$

$(2x+3)(x-2)=0$

$\implies\,x=\dfrac{-3}{2},2$

$\therefore\textbf{Roots of f(x) are \bf\,-1,\dfrac{-3}{2},2 }$

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Is the roots of 32 x cube minus 48 x square + 22 x minus 3 equal to zero are in AP then the middle root is

https://brainly.in/question/17305036

Answer 2

f(x) =2x^3+x^2-5x-2x-6 2x^3-2x+x^2-5x-6

2x (x^2-1) +x^2-6x+x-6 2x(x+1) (x-1) +x(x-6) +1(x-6) 2x(x+1) (x-1) +(x-6) (x+1) (x+1) (2x(x-1) +x-6) (x+1) (2x^2-2x+x-6) (x+1) (2x^2-x-6) (x+1) (2x^2-4x+3x-6) (x+1) (2x(x-2) +3(x-2)) f(x) =(x+1) (x-2) (2x+3) root of polynomial in given by (x+1), (x-2), (2x+3) =0 (x=-1, x=2, x=-3/2)