Family of polynomials

Question

Find two unique functions that satisfy the conditions
P(-4) = P(-3/4) = P(1/2) = 0 and p(-2) = 60

Family of polynomials 

Answer 1

A cubic polynomial is uniquely specified by 4 points. Since your function has zeros at -4, -3/4, and 1/2, this cubic function must be of the form P(x) = a(x+4)(x+3/4)(x-1/2), where the constant a follows from the fourth point: P(-2)=a(2)(-5/4)(-5/2)=60, which gives us a=48/5. Therefore,

P(x)=48/5 (x+4)(x+3/4)(x-1/2).

 

A quartic polynomial is not uniquely specified by 4 points; however, a homogeneous quartic polynomial (one that passes through the origin) is: P(x) = bx(x+4)(x+3/4)(x-1/2), where the constant b follows again from the fourth point: P(-2)=b(-2)(2)(-5/4)(-5/2)=60, which gives us b=-24/5. Therefore,
P(x)=-24/5 x(x+4)(x+3/4)(x-1/2).

Answer 2

Hi Lawson;

(P+4)(P+3/4)(P-1/2)=0

Let's FOIL the first two parenthetical equations...

FIRST...(P)(P)=P²

OUTER...(P)(3/4)=(3/4)P

INNER...(4)(P)=4P

LAST...(4)(3/4)=3

P²+(3/4)P+4P+3

[P²+(4 3/4)P+3](P-1/2)

Now, let's multiply P of the second parenthetical equation by each item of the bracketed equation...

P³+(4 3/4)P²+3P

Next, let's multiply -1/2 of the second parenthetical equation by each item of the bracketed equation...

(-1/2)P²-(2 3/8)P-(1 1/2)

Let's add the two results together...

P³+(4 1/4)P²+(5/8)P-(1 1/2)=0

Let's multiply both sides by 8 to eliminate annoying fractions...

8P³+34P²+5P-12=0

 

PluggingP=-2 does not produce the result of 60.