The real zeros of the polynomial P(x)=-x^3+15x^2-66x+80?

Question

can someone please show me how to get the "real zeros"??

Answer 1

Hi,

This equation only has 1 real zero as it only crosses the x axis once. If you enter the equation into your calculator and set your window to something close to:
xmin=-25
xmax = 5
xscl=0
ymin=-40
ymax=135
yscl=0

you will see the graph crosses over the x axis in the negative x values and then bends down on the right side before turning back up toward infinity, without intersecting the x axis again.

To find the one x intercept on th e TI83, press [2nd][TRACE] to get to the CALC commands. Choose 2: zero. Select a left bound of - 25 and a right bound of 0 and for guess, just press ENTER. The calculator will find your only real root to be x = -18.74798.

Because of the graph bending and turning away from the x axis, the other 2 roots are complex with "i".

I hope that helps!! :-)

Answer 2

well to start off, you know there will either be 1 or 3 real zeros since imaginary solutions will occur in conjugate pairs since the polynomial coefficients are all real.

first try to factor

(x + d)*(ax^2 + bx + c) = ax^3 + (ad + b)x^2 + (c + db)x + (dc) = P(X)

so we know a = -1 and d*c = 80. this tells us that d is a number that divides 80

d = 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

try d = 1 => try x = 1, P(1) = 28
try d = 2 => try x = 2, P(2) = 0!

so now factor out (x - 2) from P(x)

2 | -1 15 -66 80
...........-2 +26 -80
.....-1 + 13 -40 0

so the resulting polynomial is

P(x) = (x - 2)*Q(x)
Q(x) = -x^2 + 13x - 40
now factor this or use quadratic equation

40 = 8*5 and 13= 8 + 5

so Q(x) = -(x - 8)*(x - 5)
so P(x) = -(x - 2)(x - 5)(x - 8)

so the zeros are x = 2,5,8. If you only had 1 real solution, you could still use this approach but you'd find that the resulting Q(x) would not have real solutions!

Answer 3

2,5,8
You should try a few small numbers like 0,1,2 to see if any of them make f(x) = 0. I found x=2 makes f(x) = 0 so x-2 is a factor. Now use long division to find the quadratic factor and you will find that it can be factored to (x-5)(x-8)