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4vir4ik [10]
2 years ago
9

Which polynomial function has a leading coefficient of 1 and roots (7 i) and (5 – i) with multiplicity 1? f(x) = (x 7)(x – i)(x

5)(x i) f(x) = (x – 7)(x – i)(x – 5)(x i) f(x) = (x – (7 – i))(x – (5 i))(x – (7 i))(x – (5 – i)) f(x) = (x (7 – i))(x (5 i))(x (7 i))(x (5 – i))
Mathematics
1 answer:
Nuetrik [128]2 years ago
5 0

It looks like you might have intended to say the roots are 7 + i and 5 - i, judging by the extra space between 7 and i.

The simplest polynomial with these characteristics would be

f(x) = (x - (7 + i)) (x - (5 - i))

but seeing as each of the options appears to be a quartic polynomial, I suspect f(x) is also supposed to have only real coefficients. In that case, we need to pair up any complex root with its conjugate to "complete" f(x). We end up with

f(x) = (x - (7 + i)) (x - (7 - i)) (x - (5 - i)) (x - (5 + i))

which appears to most closely resemble the third option. Upon expanding, we see f(x) does indeed have real coefficients:

f(x) = (x^2 - (7^2 - i^2)) (x^2 - (5^2 - i^2)) = (x^2 - 8) (x^2 - 6) = x^4 - 14x^2 + 48

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Answer:

Please check the explanation.

Step-by-step explanation:

Given the equations

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solving the system of the equations

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\underline{y+6x=-10}

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\mathrm{For\:}y+6x=-10\mathrm{\:plug\:in\:}x=\frac{11}{3}

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