from the derivative of x^3+3x^2+x+3
2006-10-03 04:47:47 · 6 answers · asked by Ken B 1 in Science & Mathematics ➔ Mathematics
This doesn't factor nicely. Use the quadratic formula to find the roots r1 and r2. After simplifying, you should get
r1,2 = -1 +/- sqrt(6)/3
2006-10-03 04:52:36 · answer #1 · answered by James L 5 · 0⤊ 0⤋
x((x+2)(x+1))+3
2006-10-03 05:54:11 · answer #2 · answered by Toy 2 · 0⤊ 0⤋
I think you want to prove derivative of
x^3+3x^2+x+3
derivative of x^n = nx^n-1
so d/dx(x^3+3x^2+x+3)
= d/dx(x^3) + 3d/dx(x^2) + d/dx(x) +d/dx(3)
= 3x^2 + 3.2x + 1 + 0
= 3x^2 + 6x +1
factor
3x^2+6x+1
2006-10-03 04:55:39 · answer #3 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
(-6+/-sqrt(36-12))/6
-1+/-sqrt(24)/6
-1+4sqrt(6)/6 = -1+2*sqrt(6)/3
-1-4sqrt(6)/6 = -1-2*sqrt(6)/3
2006-10-03 04:55:12 · answer #4 · answered by bob h 3 · 0⤊ 0⤋
(3x+1)(x+1)
2006-10-03 04:50:45 · answer #5 · answered by flounder_bob@sbcglobal.net 2 · 0⤊ 2⤋
yikes!!! the most difficult subject i ever dealt with.
2006-10-03 04:55:30 · answer #6 · answered by Mayank Maheshwari 2 · 0⤊ 1⤋