Find polynomial function with a degree of 3, P(2)=5 and the given zeros of -3,-1 and 4

Answer 1

If you are told the zeros then you can convert those to factors. Remember that to find zeros, you set a polynomial equal to zero, factor it, and set each factor equal to zero. The zero "4" converts to the factor (x - 4) because when you set (x - 4) equal to zero, you get x = 4.

So the factors become (x + 3)(x + 1)(x - 4) making the polynomial P(x) = (x + 3)(x + 1)(x - 4). But there can also be a multiplier (we'll call it "a") in front of the polynomial so it can actually be P(x) = a(x + 3)(x + 1)(x - 4). To find out what "a" is, we can plug in the data point they provided (2, 5) as an x and y and solve for "a". That means we have:

5 = a(2 + 3)(2 + 1)(2 - 4)

5 = a(5)(3)(-2)

5 = -30a

a = -5/30 = -1/6

So the polynomial is:

P(x) = -1/6(x + 3)(x + 1)(x - 4)

This is in factored form and the problem doesn't say that a particular form is needed so that should be sufficient; however, if you wanted to put it in standard form, you would multiply it out:

(x + 3)(x + 1) = x² + 4x + 3 and (x² + 4x + 3)(x - 4) = x³ - 4x² + 4x² - 16x + 3x - 12 or x³ - 13x - 12 then multiplying -1/6(x³ - 13x - 12) gives P(x) = -1/6x³ + 13/6x + 2

Answer 2

If you multiply the equation for the roots y = (x+3)(x+1)(x-4) you get x³ -13x -12

Substituting 2 into the polynomial you get -30. So, in order to obtain 5, you must multiply the polynomial by -1/6.

Therefore, the final polynomial will be y = (-1/6)(x³ -13x -12)