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How do you factor the expression x squared plus three?
x^2 + 3 = x^2 - (-3) = x^2 - 3i^2 = (x - i√3)(x + i√3)
What is 3i to the 5th power?
(3i)5 = 35 x i5 = 243 x i2 x i2 x i = 243 x (-1) x (-1) x i = 243i
What is the cube root for 27?
The cube root of 27 is 3. 27 has three cube roots, a real cube root (3) and two imaginary cube roots. To compute them, start with the following equation and solve for x: x^3 = 27 x^3 - 27 = 0 Factor using the difference of cubes: (x - 3)(x² + 3x + 9) = 0 One root is x = 3. The other two can be found using the quadratic formula on: x² + 3x + 9 = 0 a = 1 b = 3 c = 9 x = [-3 ± √(9 - 4*1*9) ] / 2 x = (-3 ± √(-27) ] / 2 x = (-3 ± 3i√3 ) / 2 x = -3/2 ± (3i√3)/2 Answers: 3 -3/2 + (3i√3)/2 -3/2 - (3i√3)/2 Hope I helped :)
Can you find a third degree polynomial equation with rational coefficients that has the given numbers as roots 3i and 7?
Yes, easily. Even though the question did not ask what the polynomial was, only if I could find it, here is how you would find the polynomial: Since the coefficients are rational, the complex (or imaginary) roots must form a conjugate pair. That is to say, the two complex roots are + 3i and -3i. The third root is 7. So the polynomial, in factorised form, is (x - 3i)(x + 3i)(x - 7) = (x2 + 9)(x - 7) = x3 - 7x2 + 9x - 63
What is a polynomial with a degree of 3 and zeros of -3 and 3-2i?
The question has a simple answer only if the polynomial has rational coefficients. However, the question does not state that it has rational coefficients so it is not valid to assume that is the case. So, suppose the third root is p. Then the polynomial is (x + 3)*(x - 3 + 2i)*(x - p) = (x2 + 2xi - 9 + 6i)*(x - p) = x3 + (2i - p)x2 - (2pi - 6i + 9)x + (9p + 6pi)
What are factors of 576 that add up to -12?
564
Solve by factoring 3x2 equals -9?
3x2 = -9 (divide both sides by 3) x2 = -3 (x would have to be the square root of -3) x = ±√-3 x = ±√3i Since you want to solve by factoring: x2 = -3 add 3 to both sides x2 + 3 = 0 x2 - 3i2 = 0 x2 - (√3i)2 = 0 Factor: (x - √3i)(x + √3i) = 0 x - √3i = 0 or x + √3i = 0 x = √3i or x = -√3i
How do you factor and equation 2X2-6X plus 9?
The best way to factor this is to use the quadratic equation. Using an equation of the form: ax2 + bx + c = 0 You use the quadratic formula like this: x = [-b +(or -) sqrt(b2-4ac)]/2a For your problem, a = 2, b = -6, and c = 9. Therefore: x = [6 +(or -) sqrt(36 - 4(2)(9))]/4 = [6 +(or -) sqrt(36-72)]/4 = [6 +(or -) sqrt(-36)]/4 = (6 +(or -)6i)/4 = 3/2 +(or -)3i/2 Therefore the factorization is (x - 3/2 - 3i/2)(x - 3/2 + 3i/2).
How do you get the solution to 6 divided by the square root of -10?
16
How do you factor x squared plus 25?
(x-5i)(x+5i)
What is the factor of x2 plus 9?
x2 + 9 has no real factors. Its complex factors are (x + 3i) and (x - 3i) where i is the imaginary square root of -1.
Is -6i a complex number?
A complex number is x+iy and -6i is 0-6i or 0-i6 I would then answer yes.
How do you factor the expression x squared plus three?
x^2 + 3 = x^2 - (-3) = x^2 - 3i^2 = (x - i√3)(x + i√3)
Can x squared plus 25 be factored?
Not in real numbers. It can be factored to (x - 5i)(x + 5i) where i is the square root of negative one.
What are the factors of x to the power 6 plus nine?
x6 + 9= x6 - (-9) since i2 = -1= (x3)2 - 9i2 factor the difference of two squares= (x3 + 3i)(x3 - 3i) since 3 = (31/3)3 and -i = i3 we can write:= [x3 - (31/3)3i3] [x3 + (31/3)3i3]= [x3 - (31/3i)3] [x3 + (31/3i)3] factor the sum and the difference of two cubes= [(x - 31/3i)(x2 + 31/3ix + (31/3)2i2)] [(x + 31/3i)(x2 - 31/3ix + (31/3)2i2)]= [(x - 31/3i)(x2 + 31/3ix - (31/3)2)][(x + 31/3i)(x2 - 31/3ix - (31/3)2)]Thus, we have two factors (x - 31/3i) and (x + 31/3i),so let's find four othersAdd and subtract x2/4 to both trinomials[x2 - x2/4 + (x/2)2 + 31/3ix - (31/3)2] [x2 - x2/4 + (x/2)2 - 31/3ix - (32/3)2] combine and factor -1= {3x2/4 - [((x/2)i))2 - 31/3ix + (31/3)2]}{3x2/4 - [((x/2)i))2 + 31/3ix + (32/3)2]} write the difference of the two squares= {((3)1/2x/2))2 - [(x/2)i - 31/3]2}{((3)1/2x/2))2 - [(x/2)i + 32/3]2]} factor the difference of two squares= {[(31/2/2)x - ((1/2)i)x - 31/3)] [((31/2/2)x + ((1/2)i)x - 31/3)]} {[((31/2/2)x) - (((1/2)i)x + 31/3)] [((31/2/2)x) + ((1/2)i)x + 31/3)]}= {[(31/2/2)x - ((1/2)i)x + 31/3)] [((31/2/2)x + ((1/2)i)x - 31/3)]} {[((31/2/2)x) - ((1/2)i)x - 31/3)] [((31/2/2)x) + ((1/2)i)x + 31/3)]} simplify= {[((31/2 - i)/2))x + 31/3)] [((31/2 + i)/2))x - 31/3)]} {[((31/2 - i)/2))x - 31/3)] [((31/2+ i)/2))x + 31/3)]}so we have the 6 linear factors of x2 + 9.1) (x - 31/3i)2) (x + 31/3i)3) [((31/2 - i)/2))x + 31/3)]4) [((31/2 + i)/2))x - 31/3)]5) [((31/2 - i)/2))x - 31/3)]6) [((31/2+ i)/2))x + 31/3)]Check: Multiply:[(1)(2)][(3)(5)][(4)(6)]A) (x - 31/3i)(x + 31/3i) = x +(31/3)2B) [((31/2 - i)/2))x + 31/3)] [((31/2 - i)/2))x - 31/3)] = [(1 - (31/2)i)/2]x2 - (31/3)2C) [((31/2 + i)/2))x - 31/3)][((31/2+ i)/2))x + 31/3)] = [(1 + (31/2)i)/2]x2 - (31/3)2Multiply B) and C) and you'll get x4 - (31/3)2x2 + (31/3)4Now you have:[x +(31/3)2][x4 - (31/3)2x2 + (31/3)4] = x6 + 9
What is X when raised to the 3-5i power and the answer is 23-14i?
X = 1.31356+0.612045*iSteps to solve, take the natural log of both sides:ln(X^(3-5i)) = ln(23-14i).(3-5i)*ln(X) = ln(23-14i). Convert 23-14i to exponential form: A*e^(iΘ) {A = 26.926 and Θ = -0.54679 radians}(3-5i)*ln(X) = ln(A*e^(iΘ))= ln(A) + iΘ = ln(26.926) - 0.54679i.divide by (3-5i): ln(X) = (ln(A) + iΘ) / (3-5i) = (3.2931 - 0.54679i)/(3-5i)So we have ln(X) = 0.370978 + 0.436033i, then:e^(ln(X)) = e^(0.370978 + 0.436033i) --> X = 1.31356+0.612045*i
How do you find a 4th degree polynomial with zeros x equals 4 x equals -3 and x equals 6i?
There are only three roots given so, in general, there is no unique answer. However, if it is a real polynomial, then its complex roots must come in conjugate pairs. Then 6i is a root implies that -6i is a root. So the polynomial is (x - 4)(x + 3)(x + 6i)(x - 6i) = (x2 - x - 12)(x2 + 36) = x4 + 36x2 - x3 - 36x - 12x2 - 432 = x4 - x3 + 24x2 - 36x - 432
