Find the zeros (roots) of the following equations.
1 answer:
9514 1404 393
Answer:
x ∈ {1/2, 1, 3, i, -i}
Step-by-step explanation:
We assume your function is supposed to be ...
f(x) = 2x^5 -9x^4 +12x^3 -12x^2 +10x -3=0
I find it convenient to use a graphing calculator to find the real roots of higher-degree polynomials. The attached shows the real roots are ...
x = 1/2, 1, 3
Factoring those out leaves the quadratic factor ...
x^2 +1 = 0
That one has roots ±i.
So, the full complement of roots is ...
x ∈ {1/2, 1, 3, i, -i}
_____
Descartes' rule of signs tells you the real roots are all positive. The rational root theorem tells you that rational roots are in the set {1/2, 1, 3/2, 3}. The sum of coefficients is 0, telling you that x=1 is a real root.
Factoring that from the polynomial leaves the 4th-degree factor ...
2x^4 −7x^3 +5x^2 −7x +3 . . . . . division shown in 2nd attachment
For x=1, this has a negative value (-4), so you know there is a root between f(0) =3 and f(1) = -4. The only rational choice is 1/2.
Similarly, factoring that from the polynomial (along with a factor of 2) leaves the cubic ...
x^3 -3x^2 +x -3 . . . . . division shown in 3rd attachment
This is clearly factorable by grouping to (x -3)(x^2 +1), so we have our third real root and the quadratic that yields the complex roots.
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Answer:
B
Step-by-step explanation:
Given,
Dividend =
Divisor =
Step 1: First The dividend is and Divisor is
when divided first time the quotient will be
and remainder will be
Step: 2 Now the new dividend is and Divisor is
when divided the quotient will be
and remainder will be
Step: 3 Now the new dividend is and Divisor is
when divided the quotient will be
and remainder will be
Step: 4 Now the new dividend is and Divisor is
when divided the quotient will be
and remainder will be
Step: 5 Now the new dividend is and Divisor is
when divided the quotient will be
and remainder will be 0.
Hence When the polynomial is divided by
the answer or quotient will be equal to
and remainder will be 0.
