Linear Factorization of ${x}^{4}+{x}^{3}+5x-10$

cistG
2021-03-06
Answered

Linear Factorization of ${x}^{4}+{x}^{3}+5x-10$

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timbalemX

Answered 2021-03-07
Author has **108** answers

Step 1

Given quartic equation is :

${x}^{4}+{x}^{3}+5x-10=0$

Let us first take x common,

${x}^{4}+{x}^{3}+5x-10=0$

$\Rightarrow x({x}^{3}+{x}^{2}+5)-10=0$

$\Rightarrow x({x}^{3}+{x}^{2}+5)=10$

$\Rightarrow x=10,({x}^{3}+{x}^{2}+5)=10$

Step 2

Now,

${x}^{3}+{x}^{2}+5=10$

$\Rightarrow {x}^{2}(x+1)=5$

$\Rightarrow {x}^{2}=5,(x+1)=5$

Again,

${x}^{2}=5$

$\Rightarrow x=\pm 5i$

and , x+1=5

$\Rightarrow x=4$

Step 3

The roots are :$10,4,\pm 5i$

Given quartic equation is :

Let us first take x common,

Step 2

Now,

Again,

and , x+1=5

Step 3

The roots are :

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