Y equals 20x3-15x plus 6x-2

Answer 1

y = 20x3 - 15x + 6x - 2 or
y = 20x3 - 9x - 2

Plot the graph of y = 20x3 - 9x - 2 in the window [-1, 1] by [-2, 1] in your graphing calculator..

Use 2ND/CALC/zero to find the x-intercept values which are the zeros of the function such as

x ≈ 0.76
x ≈ -0.26
x ≈ -0.5

If the equation is y = 20x3 - 15x2 + 6x - 2, then from the Descartes' rule of sign we have:

f(x) = 20x3 - 15x2 + 6x - 2 (two changes in sign, so there are two or one positive real zeros)
f(-x) = -20x3 - 15x2 - 6x - 2 (no changes in sign, there is any negative real zero)
Since there are three zeros (degree of 3), we must have only one real positive zero, and two conjugate complex zeros.

When you plot the graph of y = 20x3 - 15x2 + 6x - 2 in the same window as above, you find that the real zero is x ≈ 0.5 = 1/2.

The sum of the roots = - -15/20 = 3/4 [-(the coefficient of x2/the coefficient of x3]
The product of the roots = - -2/20 = 1/10 [odd degree -(the constant//the coefficient of x3]

Let the complex roots be a + bi and a - bi, then we have:

1/2 + a + bi + a - bi = 3/4
2a = 3/4 - 1/2
2a = 1/4
a = 1/8 = 0.125

(1/2)(a + bi)(a - bi) = 1/10
(1/2)(a2 - b2i2) = 1/10 (substitute -1 for i2)
a2 + b2 = 1/10 ÷ 1/2 (substitute 1/8 for a)
(1/8)2 + b2 = 1/10 x 2/1
1/64 + b2 = 1/5
b2 = 1/5 - 1/64
b2 = 1x64/5x64 - 1x5/64x5 = 64/320 - 5/320 = 59/320
b = ± √(59/320) ≈ ± 0.4

So that a + bi = 0.125 + 0.4i, a - bi = 0.125 - 0.4i

Thus the zeros are 0.5, 0.125 + 0.4i, and 0.125 - 0.4i.