Question

Find the polynomial function in standard form that has the zeros listed. 1(multiplicity 2), -2(multiplicity 3)​

Answer 1

Answer:

  f(x) = x⁵ +4x⁴ +x³ -10x² -4x+8

Step-by-step explanation:

If p is a zero, then (x -p) is a factor. The multiplicity of the zero tells you how many times that is a factor (its exponent).

  f(x) = (x -1)²(x +2)³ = (x² -2x +1)(x³ +6x² +12x +8)

  f(x) = x⁵ +4x⁴ +x³ -10x² -4x+8

_____

Additional comment

You can use the distributive property to write out the 12 terms of the product of the two expanded factors, or you can do a little mental arithmetic based on what you know about the exponent of a product term. (It is the sum of the exponents of its factors.)

The mental exercise can be made easier by writing the coefficients of the factors as though they were two cubics. It works well to write them in two rows:

  $\left[\begin{array}{cccc}0&1&-2&1\\1&6&12&8\end{array}\right]$

The columns are in order of decreasing powers of x. A relatively simple and symmetrical pattern of sums of cross products is used to find the coefficients of the final polynomial. Working in order of decreasing exponents, we have ...

coefficient of x^6 = (0)(1) = 0

coefficient of x^5 = (0)(6) +(1)(1) = 1

coefficient of x^4 = (0)(12) +(1)(-2) +(1)(6) = 4

coefficient of x^3 = (0)(8) +(1)(1) +(1)(12) +(6)(-2) = 1

coefficient of x^2 = (1)(8) +(6)(1) +(-2)(12) = -10

coefficient of x^1 = (-2)(8) +(12)(1) = -4

coefficient of x^0 = (1)(8) = 8