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))
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Complex roots come in conjugate pairs, meaning if (7 + i) is a root, then (7 - i) is also a root, and if (5 - i) is a root, then (5 + i) is also a root.
So the four roots are (7 + i), (7 - i), (5 - i), and (5 + i). Subtract each of them from x and multiply the factors to get the polynomial
So im thinking the third choice fits your description. because you have (x-(7-i))
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rather than just 7-x
Would you be able to help me with a similar one
If a polynomial function f(x) has roots –9 and 7 – i, what must be a factor of f(x)?
you would do the same thing right? like (x + (7 + i)) ?
yeah, but it's always "x - ___" so (x - (7 + i))
Its x- for every problem like that?
yes. even for real roots. You'd represent the root of -9 as (x - (-9)) or (x + 9)
Which second degree polynomial function has a leading coefficient of –1 and root 4 with multiplicity 2?
f(x) = –x^2 – 8x – 16
f(x) = –x^2 + 8x – 16
f(x) = –x^2 – 8x + 16
f(x) = –x^2 + 8x + 16
how do you find the multiplicity?
Multiplicity is the exponent on the factor. A root of 4 means the factor is (x - 4). Multiplicity of 2 means that (x - 4) is a factor twice and it should be written as (x - 4)². A leading coefficient of -1 means the function is y = -(x - 4)².
I'd multiply/expand -(x - 4)² to see which choice it matches