find all roots of the polynomial x3-x2+16x-16?

Answer 1

Factor by grouping:

(x^3 - x^2) + (16x - 16)
x^2 (x - 1) + 16 (x - 1)
(x^2 + 16)(x - 1)

So, we have two equations to set equal to zero:

(x - 1) = 0
==> x = 1

(x^2 + 16) = 0
x^2 = -16

IF YOU ARE ONLY LOOKING FOR REAL ROOTS, YOU SEE THAT THERE IS NO ANSWER BECAUSE NO REAL NUMBER SQUARED IS A NEGATIVE NUMBER. HOWEVER, IF YOU ARE LOOKING FOR COMPLEX ROOTS AS WELL (which the problem implies), TAKE THE SQUARE ROOT OF BOTH SIDES:

x^2 = -16
x^2 = 16 * -1
x = √16 * √(-1)
x = ± 4i

So, your roots are x = 1, x = -4i, and x = 4i
________________________________

Answer 2

The x-intercepts are x-values anyplace the function equals 0. that's many times simplest to apply trial and mistake to locate one x-intercept (or "root") then factorise the polynomial by dividing by (x-a) (the position a is the linked fee of the inspiration), and locate roots for the perfect polynomial. To get you began, try putting x=3 into h(x). you should locate x-3 is a aspect of the polynomial, so divide by (x-3) to detect a clean quadratic. And, you should use the quadratic equation or ending up the sq. to locate the roots of a quadratic very truly.

Answer 3

Graphing the equation, you find that it crosses the x-axis at only one spot - (1,0). So the root is 1.

Answer 4

x^2(x-1)+16(x-1)=(x-1)(x^2+16)
thr roots are:
1,+4i,-4i.

Answer 5

first combine all the like terms:

x-16x-16 (3x - 2x = x)

then factor, set x qual to zero and you have your zeros.

Answer 6

(x - 1), (x^2 + 16)

You can factor (x^2 + 16) to (x + 4i)(x - 4i) if you wish.