2x^3-3x^2-36x+5
2007-11-30 17:58:37 · 4 answers · asked by 2789 2 in Science & Mathematics ➔ Mathematics
I suppose y=2x^3-3x^2-36x+5
x intercept so y=0
2x^3-3x^2-36x+5=0
(x-5)(2x^2 +7x -1)=0
x=5
or x= (-7 +/- √57)/4
2007-11-30 18:13:58 · answer #1 · answered by mbdwy 5 · 0⤊ 0⤋
Interpreting x intercepts as roots of this equation.
We have 2x^3-3x^2-36x+5=0
Use scientific calculator or by trial and error, you can get one of the roots of this equation as +5.
Divide this equation by (x-5) to get (2x^2 + 7x -1) as other factor.
Solve this by formula of quadratic equation (ax^2+bx+c=0).
Roots are [-b +or- sqrt(b^2-4ac)]/2a.
So 3 roots of above equation are:
+5, -3.637, 0.137.
Thank you.
2007-12-01 02:14:45 · answer #2 · answered by Anonymous · 0⤊ 0⤋
well if it is y=your equation just plug in 0 for y because that is when it crossed the x line and then solve for x
0=2x^3-3x^2-36x+5
then your answers will be where it crosses the x axis
2007-12-01 02:14:24 · answer #3 · answered by Briana L 4 · 0⤊ 0⤋
You forgot something. That's not an equation.
2007-12-01 02:04:04 · answer #4 · answered by Demiurge42 7 · 1⤊ 0⤋