Form a third degree polynomial function with real coefficients such that 8+i and 7 are zeros.

1 answer:

3 0

The zeros of the equation are <span>8+i and 7. however since this is a third degree polynomial, the other root missing is the conjugate of the imaginary term 8-i, which is 8 +i. We multiply then ( x- 8 - i) *( x- 7)* (x - 8 + i).The expansion is (x2 - 16x + 1 --1 ) *(x-7) = </span><span>(x2 - 16x + 2 )(x-7) = x3 -7 x2 -16x2 +112 x +2x -14 = x3 -23 x2 + 114 x -14</span>

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Step by Step Explanation:

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