Question 723322: Hi what are the zeros of these functions? F(x)=3x^2+x-2 F(x)=x^3+6x^2+16 F(x)=x^3+8x^2-20x F(x)=x^3-7x^2+7x+15
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Your first function is
(1) 3x^2 + x - 2 = 0, factors into
(2) (3x - 2)(x + 1) = 0, set the factors equal to zero and solve for the root x give
(3) 3x - 2 = 0 or
(4) x = 2/3 and
(5) x + 1 = 0 gives us
(6) x = -1
Let's check the roots using (1).
Is (3*(2/3)^2 + 2/3 -2 =0)?
Is (4/3 + 2/3 -2 = 0)?
Is (6/3 - 2 = 0)?
Is (2 - 2 = 0)? Yes
Is (3(-1)^2 - 1 - 2 = 0)?
Is (3 - 3 = 0)? Yes
The roots of (1) are {-1,2/3}
Your second function is
(7) x^3 + 6x^2 + 16 = 0
Because (7) is an odd order polynomial, it must have at least one real root. You can find it by trial and check. I found the real root of (7) to approximately
(8) x = -6.391648
I don't know if this can be derived analytically or not. To find the remaining quadratic, divide (7) by (x+6.39...) and get
(9) (x^2 - (.39...)x + 2.5032...)
The quadratic (9) has a complex pair of roots at approximately
(10) x = 0.2 +or- 1.59i
You'll have to check them - way too tired tonight. I know the real root is correct. If you have a graphing calculator, graph (7), it's really neat. It's derivatives are at x = zero and x = -4. The graph show how the curve of (7) changes "direction", slope equals zero, at -4 and 0.
Your third function is
(10) x^3 + 8x^2 - 20x = 0 which factors into
(11) x(x + 10)(x - 2) = 0 giving three real roots
(12) x = {0,-10,2}
You can easily show that each value of in (12) satisfies (10).
Your fourth function is
(13) x^3 - 7x^2 + 7x + 15 = 0
It also must have one real root so let's find it. I start by letting the sum of last two terms equal to zero or
(14) 7x + 15 = 0 or
(15) x = -15/7 or x is about -2. Let's try x = -2.
Is ((-2)^3 - 7(-2)^2 + 7(-2) + 15 = 0)?
Is (-8 - 28 -14 +15 = 0)?
Is (-35 = o)? not even close! But we know that the rel root is between -2 and 0, because f(-2)<0 and f(0)=15>0. So let's try x = -1
Is ((-1)^3 - 7(-1)^2 + 7(-1) + 15 = 0)?
Is (-1 -7 -7 +15 = 0)? Yes, we found the real root!!
(16) x = -1
Divide (13) by (x+1) to find the quadratic, and get
(17) x^2 -8x +15 = 0 which factors to
(18) (x-5)(x-3) = 0 and the roots are
(19) x = {5,3}, and for your fourth function (13) the roots are
(20) x = (-1,3,5}
PS My roots are your zeroes.
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