Question

Using the linear factorization theorem to find a polynomial with given zeros. Find a fourth degree polynomial with real coefficients that has zeros of -3,2,i, such that f( -2) =100

Answer 1

The fourth-degree polynomial with real coefficients, having zeros of -3, 2, i, and satisfying f(-2) = 100, is given by the equation
`\( f(x) = (x + 3)(x - 2)(x - i)(x + i) \).`

Step-by-step explanation:

To find a polynomial with given zeros, we use the linear factorization theorem, which states that for every zero (a), there is a factor
`\((x - a)\)` in the polynomial. Given the zeros -3, 2, i, we form the linear factors
`\((x + 3)(x - 2)(x - i)(x + i)\)`. Multiplying these factors gives the desired fourth-degree polynomial.

Now, to ensure the polynomial has real coefficients, we need to consider the conjugate pairs of complex zeros. Since (i) is a zero, its conjugate (-i) is also a zero. Therefore, the expression becomes
`\( (x + 3)(x - 2)(x^2 + 1) \).`

To satisfy
`\(f(-2) = 100\)`, we substitute (x = -2) into the polynomial:

`[ f(-2) = (-2 + 3)(-2 - 2)((-2)^2 + 1) = 1 \cdot (-4) \cdot 5 = -20 \]`

The given condition is not met with the current form of the polynomial. To adjust for this, we can multiply the polynomial by (-5) to get
`\( f(x) = -5(x + 3)(x - 2)(x^2 + 1) \)`. Now,
`\(f(-2) = 100\)`, and the polynomial satisfies all given conditions.