1. State how many imaginary and real zeros the function has. f(x) = x^3 + 5x^2 – 28x – 32?

2. Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.

8, -14, and 3 + 9i

2 Answers

  • 1) This f(x) has at least one real zero; a quick trial and error should tell you x = -1 is a zero. Dividing f(x) by x + 1 you got x^2 + 4x – 32, which becomes (x – 4)(x + 8); x = 4 and x = -8 are therefore zeros. You have three real zeros all told.

    2) Zeros of 8 and -14 correspond to factors of x – 8 and x + 14 respectively; their product is x^2 + 6x – 112. A zero of 3 + 9i comes with one of 3 – 9i, as complex zeros come as conjugate pairs when trying to construct a polynomial with real number coefficients. Their product is:

    (x – (3 + 9i))(x – (3 – 9i)) = x^2 – 3x + 9ix – 3x + 9 – 27i – 9ix + 27i – 81i^2 = x^2 – 6x + 90.

    (x^2 + 6x – 112)(x^2 – 6x + 90) = x^4 – 6x^3 + 90x^2 + 6x^3 – 36x^2 + 540x – 112x^2 + 672x – 10080…

    … f(x) = x^4 – 58x^2 + 1212x – 10080.

  • Calm down now man k

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