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Q: What is the polynomial roots to 1 5i and -3i?
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Related questions

How do you subtract imaginary numbers?

When adding and subtracting complex numbers, you can treat the "i" as any variable. For example, 5i + 3i = 8i, 5i -3i = 2i, etc.; (2 + 5i) - (3 - 3i) = (2 - 3) + (5 + 3)i = -1 + 8i.


Which polynomial has rational coefficients a leading leading coefficient of 1 and the zeros at 2-3i and 4?

There cannot be such a polynomial. If a polynomial has rational coefficients, then any complex roots must come in conjugate pairs. In this case the conjugate for 2-3i is not a root. Consequently, either (a) the function is not a polynomial, or (b) it does not have rational coefficients, or (c) 2 - 3i is not a root (nor any other complex number), or (d) there are other roots that have not been mentioned. In the last case, the polynomial could have any number of additional (unlisted) roots and is therefore indeterminate.


Plot the number in a complex plane -1-3i?

2


Let A equals 1 plus 2i and let C equals 5 plus 5i what is A-C?

-4-3i


What is the answer if you Find all the roots of the equation x4 - 2x3 plus 14x2 - 18x plus 45 0 given that 1 plus 2i is one of its roots.?

The four roots are:1 + 2i, 1 - 2i, 3i and -3i.


What is the minimum number of roots an odd powered polynomial will have?

1


How many unique roots will a third degree polynomial function have?

It can have 1, 2 or 3 unique roots.


What is 1 divided by 2 plus 5i in the form a plus bi?

1/(2 + 5i) (multiply both the numerator and the denominator by 2 - 5i)= 1(2 - 5i)/(2 + 5i)(2 - 5i)= (2 - 5i)/(4 - 25i2) (substitute -1 for i2)= (2 - 5i)/(4 + 25)= (2 - 5i)/29= 2/29 - (5/29)i


What is the reciprocal of 3 plus 5i?

1/(3+5i)=(3-5i)/((3+5i)(3-5i))=(3-5i)/(9+25)=(3-5i)/34


What are some imaginary numbers?

i is the Imaginary Unit, equal to sqrt(-1). So i and any real number multiplied by i will all be imaginary numbers. Here are some: i, -i, 5i, -3i, i*pi, etc.


What are the roots of the polynomial function 3x2-13x 4?

(3x - 1)(x - 4) so roots are 1/3 and 4


What are the three cube roots of negative 8?

The 3 cube roots of a cubic equation are often going to require complex numbers, and this is one of them. The primary root is (-2) since (-2)3= -8 The other two roots are going to have the same mod (2) but be evenly spaced around the complex plane, so (2pi)/3 angular difference.These angles are pi/3 and 5pi/3(2,pi/3) = (1 + √3i) = (1 + 1.732i)(2,5pi/3) = (1 + -√3i) = (1- 1.732i)Check(-2)3 = -8◄(1 + √3i)3 = -8◄(1 - √3i)3 = -8◄■