If you have a real polynomial (one where the coefficients are real numbers - rationals are part of the reals), THEN any unreal root must be accompanied by its "conjugate", a number where the unreal part has the opposite sign.
4i (an imaginary number) is the same as
(0 + 4i) a Complex number = combining a real part (0) with an imaginary part (4i).
Its conjugate is found by changing the sign of the imaginary part:
(0 - 4i) is the conjugate of (0 + 4i) -- and vice-versa (they are conjugates of each other)
Therefore, if you are told that the polynomial has rational coefficients AND that one of the roots is 4i, then there MUST be another root at -4i.
Once you have a root (let's call it "k"), then the value (x - k) must be a factor of the polynomial.
For the root +5, the factor is (x - 5)
For the root +4i, the factor is (x - 4i)
For the root - 4i, the factor is (x + 4i)
Therefore, the polynomial, expressed as a factored expression, will look like this:
K(x - 5)(x - 4i)(x + 4i)
(where K can be any scalar = a "normal number")
Let's use K=1 to make it easy
polynomial = (x - 5)(x - 4i)(x + 4i)
begin by multiplying the complex factors:
(x - 4i)(x + 4i) = x^2 - 16i^2 = x^2 + 16
Now we have
polynomial = (x - 5)(x + 16)
which you multiply together to get an expression that beings with x^3
(making it a third-degree polynomial).
If you are asked explicitly to state it as an "equation", pick a variable for the other side of the equal sign, for example "y"
1) "a" can be any arbitrary non-zero rational value if the coefficients must be rational. The simpler case would be for a = 1.
2) If the polynomial has rational coefficients, the existence of 4i as a root requires that -4i must also be a root. Complex roots of polynomials with real coefficients always occur in conjugate pairs.
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If you have a real polynomial (one where the coefficients are real numbers - rationals are part of the reals), THEN any unreal root must be accompanied by its "conjugate", a number where the unreal part has the opposite sign.
4i (an imaginary number) is the same as
(0 + 4i) a Complex number = combining a real part (0) with an imaginary part (4i).
Its conjugate is found by changing the sign of the imaginary part:
(0 - 4i) is the conjugate of (0 + 4i) -- and vice-versa (they are conjugates of each other)
Therefore, if you are told that the polynomial has rational coefficients AND that one of the roots is 4i, then there MUST be another root at -4i.
Once you have a root (let's call it "k"), then the value (x - k) must be a factor of the polynomial.
For the root +5, the factor is (x - 5)
For the root +4i, the factor is (x - 4i)
For the root - 4i, the factor is (x + 4i)
Therefore, the polynomial, expressed as a factored expression, will look like this:
K(x - 5)(x - 4i)(x + 4i)
(where K can be any scalar = a "normal number")
Let's use K=1 to make it easy
polynomial = (x - 5)(x - 4i)(x + 4i)
begin by multiplying the complex factors:
(x - 4i)(x + 4i) = x^2 - 16i^2 = x^2 + 16
Now we have
polynomial = (x - 5)(x + 16)
which you multiply together to get an expression that beings with x^3
(making it a third-degree polynomial).
If you are asked explicitly to state it as an "equation", pick a variable for the other side of the equal sign, for example "y"
y = x^3 - 5x^2 + 16x - 80
P₃(x) = a(x-5)(x-4i)(x+4i)
= a(x-5)(x²+16)
= a(x³-5x²+16x-80)
Notes:
1) "a" can be any arbitrary non-zero rational value if the coefficients must be rational. The simpler case would be for a = 1.
2) If the polynomial has rational coefficients, the existence of 4i as a root requires that -4i must also be a root. Complex roots of polynomials with real coefficients always occur in conjugate pairs.