How do I enter this into Wolfram Alpha? x^3-5x^2+2x+8=0 given that -1 is a zero of f(x)=x^3-5x^2+2x+8.?

f(x) = x³ -5x² +2x+8 =0

It is a polynomial of degree 3 (highest index of x, is 3). It means it has three factors & f(x) = 0 when any of them becomes a zero.

Let one of the factors be set to zero

(x -α) = 0,

x = α

thus making f(x)=0, ' α ' the value of the variable x & is called a zero of f(x).

Following this procedure we set out find the three factors. it was already stated that one zero is -1. So

x = -1,

x+1 = 0

& (x+1) is a factor.

Let the other factors be (x -α), (x -β) so that factorisation is

f(x) = (x+1)(x -α)(x -β).

You are reminded that when

(x -α) = 0 → at x = α, f(x)=0 &

(x -β) = 0 → at x = β, f(x)=0 .

f(x) / (x+1) = (x -α)(x -β) = 0 at x = α, β.

Now divide f(x) by (x+1), by long hand division

x+1 ) x³ -5x² +2x+8 ( x² -6x+8

. . . . .x³ + x²

. . . . ________

. . . . . . .-6x² +2x

. . . . . . .-6x² - 6x

. . . . . __________

. . . . . . . . . . . . 8x+8

. . . . . . . . . . . . 8x+8

. . . . . . . . . . . _____

. . . . . . . . . . . . . . .0

hence

f(x) / (x+1) = x² -6x+8 = (x -α)(x -β) .

Now the task to factorise to get α, β.

It is simple as its is a quadratic

x² -6x+8 = x² -4x -2x+8

= x(x -4) -2(x -4) = (x -2)(x -4).

x = 2, 4 are the α, β, the other roots.

f(x) = x³ -5x² +2x+8 = (x+1)(x -2)(x -4).

{ Why do you need WolframAlpha for this !}

(x+1)(x^2 -6x +8) = 0 ---->x = .. ( 3 soluitons)