How do you find the real solutions of the polynomial $x^{3} - 7 x^{2} = 7 - x$ $x^3-7x^2=7-x$ ?

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Answer 1 · 2017-06-16T07:45:26

$x = 7$ $x=7$

$x^{3} - 7 x^{2} = 7 - x$ $x^3-7x^2=7-x$

$\Rightarrow x^{3} - 7 x^{2} + x - 7 = 0$ $=>x^3-7x^2+x-7=0$

let $f (x) = x^{3} - 7 x^{2} + x - 7$ $" "f(x)=x^3-7x^2+x-7$

for cubics there is a least one real root.

finding this root by the factor theorem. The root(s) must be a factor of $7$ $7$ .

$f (7) = 7^{3} - 7^{2} \times 7 + 7 - 7$ $f(7)=7^3-7^2xx7+7-7$

$f(7)=cancel(7^3-7^3)+cancel(7-7)=0$

$:. x=7" is a root"$

to see if there are any other real root we factorise

$x^3-7x^2+x-7=(x-7)(x^2+bx+c)$

comparing constants

$=>c=1$

comparing coefficients of $" " x^2$

$LHS=-7$

$RHS=-7+b=>b=0$

giving

$x^3-7x^2+x-7=(x-7)(x^2+1)$

other roots $" "x^2+1=0$

no real roots.

$:.$ the only real root is $x=7$

How do you find the real solutions of the polynomial x^3-7x^2=7-xx3−7x2=7−xx^3-7x^2=7-x?