consider the polynomial f(x)=1+2x+3x^2+4x^3 ?

Welcome to my article consider the polynomial f(x)=1+2x+3x^2+4x^3 ?. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question consider the polynomial f(x)=1+2x+3x^2+4x^3 ?, read and understand it carefully till the end.

Let us know how to solve this question consider the polynomial f(x)=1+2x+3x^2+4x^3 ?

First write the question on the page of the notebook.

consider the polynomial f(x)=1+2x+3x^2+4x^3 ?

Consider the polynomial f(x) = 1 + 2x + 3×2 + 4×3. Let s be the sum of all distinct real roots of f(x) and let t = |s|

since,

\displaystyle {{f}_{{(x)}}}=1+2x+3{{x}^{2}}+4{{x}^{3}} ,

\displaystyle f{{'}_{{(x)}}}=0+2+6x+12{{x}^{2}} ,

\displaystyle f{{'}_{{(x)}}}=12{{x}^{2}}+6x+2 ,

here,

a = 12 ,

b = 6 ,

c = 2

\displaystyle D={{b}^{2}}-4ac ,

\displaystyle D={{6}^{2}}-4\times 12\times 2 ,

\displaystyle D=36-96 ,

\displaystyle D=-60 ,

that is ,

\displaystyle D=-60<0 ,

Only one real root for f(x)=0

Also f(0) =1 , f(-1) = -2

Root must lie in ( -1 , 0 )

Taking averate at 0 and -1

f( -1/2) = 1/4 implies

\displaystyle Root\mu stlie\in \left( {-1,-\frac{1}{2}} \right) ,

\displaystyle similarly\left( {-\frac{3}{4}} \right)=\frac{1}{2}implies ,

\displaystyle Root\mu stlie\in \left( {-\frac{3}{4},\frac{1}{2}} \right)ANSWER ,

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