Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros. 3-i, squ
are root of 7
1 answer:
Real coefients
if a+bi is a root then a-bi is also a root
since 3-i is a root then 3+i s also a root
also the √7, so if √7 is a root then -√7 is also a root
so
f roots are r1 and r2 then the factored form is
f(x)=(x-r1)(x-r2)
roots are 3-i, 3+i and √7 and -√7
f(x)=(x-(3-i))(x-(3+i))(x-√7)(x-(-√7))
expanded
f(x)=x⁴-6x³+3x²+42x-70
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Hope this helped.
So 5^2 -4^2 is 25-16 which is 9
6^2/9 is 36/9 which is 4 +9 is 13
13(13) is 169
Answer: 5(x-1)
Step-by-step explanation:
5*x = 5x
5*-1 = -5
therefore the answer is 5(x-1)
2.9411764706
but if they're telling you to round, then round (: