Question 391718: The zeros of f(x) = 3x^4 + 8x^3 + 6x^2 + 3x - 2 are?
I know it's 4. And I have an idea of what they are... But I'm confused. Please show the steps, if possible.
Given that 4i is a zero of f(x) = x^4 + 13x^2 - 48, there must be three real zeros of f(x).
True or false?
For this one, and if possible, could someone provide a brief explanation of what's the difference between calculating real zeros and imaginary zeros? That's the main issue I have.
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! "The zeros of f(x) = 3x^4 + 8x^3 + 6x^2 + 3x - 2 are?
I know it's 4. And I have an idea of what they are... But I'm confused. Please show the steps, if possible.
Given that 4i is a zero of f(x) = x^4 + 13x^2 - 48, there must be three real zeros of f(x).
True or false?
For this one, and if possible, could someone provide a brief explanation of what's the difference between calculating real zeros and imaginary zeros? That's the main issue I have."
f(x) = 3x^4 + 8x^3 + 6x^2 + 3x - 2,
+- 1 and +-2 are factors of -2, trying x + 2
..........3x^3 + 2x^2 + 2x - 1
x + 2 --> 3x^4 + 8x^3 + 6x^2 + 3x - 2
..........3x^3 + 6x^3
.................2x^3 + 6x^2
.................2x^3 + 4x^2
........................2x^2 + 3x
........................2x^2 + 4x
...............................-x - 2
...............................-x - 2
+-1 and +- 3 are factors of 3, and +-1 are factors of -1, trying 3x - 1
...........x^2 + x + 1
3x - 1 --> 3x^3 + 2x^2 + 2x - 1
...........3x^3 - x^2
..................3x^2 + 2x
..................3x^2 - x
.........................3x - 1
.........................3x - 1
f(x) = 3x^4 + 8x^3 + 6x^2 + 3x - 2,
we got x + 2 and 3x - 1 as factors so far
can x^2 + x + 1 be factored?
use quadratic formula, a = 1, b = 1, c = 1
x = -1/2 +- sqrt(3)i/2
f(x) = 3x^4 + 8x^3 + 6x^2 + 3x - 2
f(x) = (x + 2)(3x - 1)(x + 1/2 - sqrt(3)i/2)(x + 1/2 + sqrt(3)i/2)
x + 2 = 0 --> x = -2
3x - 1 = 0 --> 3x = 1 --> x = 1/3
x = -1/2 + sqrt(3)i/2,
the sqrt(3)i/2 is the imaginary part of a + bi or the bi
x = -1/2 - sqrt(3)i/2,
the -sqrt(3)i/2) is the imaginary part of a - bi or the -bi
f(x) = x^4 + 13x^2 - 48
f(x) = (x^2 - 3)(x^2 + 16) by FOIL same as above
x^2 - 3 = (x + sqrt(3))(x - sqrt(3))
x^2 + 16 = (x + 4i)(x - 4i) --> 4i * -4i = -16i^2 = -16 * -1 = 16, i^2 = -1
x = +- sqrt(3) --> 2 real zeroes
x = +- 4i --> 2 imaginary zeroes
false, not 3 real but 2
calculating real zeroes and imaginary zeroes is pretty much the same,
imaginary part of a complex number a+bi or a-bi (a+bi and a-bi are conjugates) is the +- bi, a and b are real numbers, sqrt(-1) = i, i^2 = -1
(a + bi)(a - bi) = a^2 - b^2 * i^2 = a^2 + b^2
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