Find a degree 3 polynomial

Question

Find a degree 3 polynomial that has zeros - 2, 3 and 7 and in which the coefficient of x^2 is -16.

Answer 1

If you want a degree three polynomial, then we must have the form:

 

f(x) = ax³ + bx² + cx + d, where a, b, c, and d are all real numbers.

 

Luckily, we are given that the solutions are -2, 3, and 7. So we can just say outright that the polynomial can be:

 

f(x) = (x+2)(x-3)(x-7) = x³ - 8x² + x +42. But the coefficient of the x² term isn't -16!

 

That's okay, because we're trying to pull a random function out of our butts, so we can kinda do whatever we want to this thing to make it fit what the problem needs.

 

If we multiply the whole thing by 2, we see that the -8x² becomes a -16x², which is what we want!

 

Multiplying everything by two is the same thing as:

 

f(x) = 2(x+2)(x-3)(x-7) = 2x³ - 16x² + 2x +84

 

Judging from the first equality, we can easily see that the zeros do not change just because we added a constant out in front. And badaboom, we have our answer!