Abbey Ellison

2022-04-05

Finding real roots of $P\left(x\right)={x}^{8}-{x}^{7}+{x}^{2}-x+15$

tralhavahr9c

Expert

we have to calculate no. of ral roots of ${x}^{8}-{x}^{7}+{x}^{2}-x+15=0$
Solution: Let $f\left(x\right)={x}^{8}-{x}^{7}+{x}^{2}-x+15$, we check real solution in $x\in \mathbb{R}$
If $x\ge 0$, then $f\left(x\right)={x}^{8}-{x}^{7}+{x}^{2}-x+15>0$
If $0, then $f\left(x\right)={x}^{8}+{x}^{2}\left(1-{x}^{5}\right)+\left(1-x\right)+14>0$
If $x\ge 1$, Then $f\left(x\right)={x}^{7}\left(x-1\right)+x\left(x-1\right)+15>0$
So we observe that $f\left(x\right)={x}^{8}-{x}^{7}+{x}^{2}-x+15>0;\mathrm{\forall }x\in \mathbb{R}$
So $f\left(x\right)={x}^{8}-{x}^{7}+{x}^{2}-x+1=0$ has no real roots for all $x\in \mathbb{R}$

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