SOLUTION: Write an equation for the third degree polynomial, (-1,0),(2,0) also f(0)=-34

Question 1135164: Write an equation for the third degree polynomial, (-1,0),(2,0) also f(0)=-34
Found 3 solutions by mathsolverplus, greenestamps, ikleyn:
Answer by mathsolverplus(88)   (Show Source): You can put this solution on YOUR website!
Given: (-1,0) (2,0) and (0,-34)
The standard form of third degree polynomial:

Substitute (0,-34):
-34 = c
Substitute (-1,0) and c=-34
0=-a+b-34
Substitute (2,0) and c=-34
0 =8a+4b-34
0=-a+b-34
0=8a+4b-34
multiply the first equation by 4,
0=-4a+4b-136
0=8a+4b-34
First equation subtracts second equation:
0=-12a-102
102=-12a
a=-17/2
b=a+34=-17/2+34=51/2
a=-17/2, b=51/2, c=-34

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Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

The solution by tutor @mathsolverplus is incomplete; the general form of a cubic equation has a linear term. His polynomial is one that passes through the three given points and has no linear term.

There are an infinite number of third degree polynomials that pass through the three given points (-1,0), (2,0) and (0,-34).

The general form of a third degree polynomial in factored form is

where p, q, and r are the three roots.

We know that two of the roots are -1 and 2, so the general form of this third degree polynomial is

The other piece of information we have is that f(0)=-34. Using that in our general form gives us

This shows us that the third root depends on the value of the leading coefficient; and that means there are an infinite number of polynomials whose graphs pass through the three given points.

The graph below shows plots of two of those functions:
(1) a=1 --> r = -17; y = (x+1)(x-2)(x+17) (red graph)
(2) a=3 --> r = -17/3; y = 3(x+1)(x-2)(x+17/3) (green graph)

Answer by ikleyn(52780)   (Show Source): You can put this solution on YOUR website!
.

The "solution" by @mathsolverplus is not only incomplete - it is incorrect.

The problem formulation is incorrect, too.

In order for the polynomial of the degree 3 could be identified / (found) by an unique way, its values should be given at 4 points,

while the condition provides its values only at 3 points.

So, both the condition and the "solution" by @mathsolverplus   ARE DEFECTIVE.

--------------

It is just, probably, couple of weeks I observe an activity of @mathsolverplus at this forum.

Regarding this activity, I can resume that for routine problems @mathsolverplus, probably, can provide valid solutions
(although not always).

But for the problems "one step away from routine" the probability he provides incorrect answers becomes dangerously high.

Under supervision and correction from other tutors in this forum, it still can be kept in some reasonable frames
(although, I personally am not very happy to clean after him . . . ).

But as a self-standing tutor at his own web-site, without supervising and corrections from other tutors he is not able
to provide error-proven (error-free) solutions on stable basis.

Keep it in your mind if/when you will consider to follow him at his web-sites.

You will be at constant risk of getting wrong solutions.

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