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2 years ago

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 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) = (x2 - 16x + 2 )(x-7) = x3 -7 x2 -16x2 +112 x +2x -14 = x3 -23 x2 + 114 x -14

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3 years ago

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Step-by-step explanation:

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1 year ago

Read 2 more answers

Convert into mixed number 81/3 and -29/2

LuckyWell [14K]

Hmmmm to convert from improper to mixed, simply do the division of both, the quotient gets upfront, the remainder as the numerator, for example,

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2 years ago

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