form a polynomial with a degree of 4 zeros 4,-3i,1 and multiplicity 2
Why did the polynomial go to the circus? Because it wanted to find some zeros and have a degree of fun! 🎪
Alright, let's create a polynomial based on your specifications. Since the zeros are 4, -3i, and 1 (with a multiplicity of 2), we can write the polynomial as:
(x - 4)(x + 3i)(x + 3i)(x - 1)
Multiplying this out, we get:
(x - 4)(x^2 + 6ix - 9)(x - 1)
Now, let's simplify further:
(x^3 - x^2 + 6ix^2 - 6ix - 9x + 9 - 4x^2 + 4x - 6ix + 6i - 9)
Consolidating like terms, we get:
x^3 - 5x^2 + 3ix^2 - 15x - 4ix + 13 - 6i
And there you have it, a polynomial with a degree of 4, zeros of 4, -3i, 1, and a multiplicity of 2! 🎉
To form a polynomial with a degree of 4 and given zeros, we can use the fact that if a complex number is a zero, then its conjugate is also a zero. Given that the zeros are 4, -3i, -3i (conjugate of +3i), and 1 (with multiplicity 2), we can create a polynomial by multiplying the factors corresponding to each zero.
Step 1: Write out the factors for each zero:
(x - 4) (corresponding to the zero 4)
(x + 3i) (corresponding to the complex zero -3i)
(x - 3i) (corresponding to the complex zero -3i's conjugate)
(x - 1)² (corresponding to the zero 1 with multiplicity 2)
Step 2: Multiply the factors:
(x - 4)(x + 3i)(x - 3i)(x - 1)²
Step 3: Simplify the expression:
(x - 4)(x² + 3ix - 3ix + 9)(x - 1)²
(x - 4)(x² + 6ix + 9)(x - 1)²
Step 4: Expand the expression:
(x - 4)(x² + 6ix + 9)(x² - 2x + 1)
(x - 4)(x⁴ - 2x³ + x² + 6ix³ - 12ix² + 6ix + 9x² - 18x + 9)
Step 5: Combine like terms:
x⁴ - 2x³ + x² + 6ix³ - 12ix² + 6ix + 9x² - 18x + 9x⁴ - 36x² + 18x + 36
10x⁴ + 4x³ - 26x² - 12x + 36
So, a polynomial with a degree of 4, zeros 4, -3i, 1 (with multiplicity 2) is 10x⁴ + 4x³ - 26x² - 12x + 36.
To form a polynomial with a degree of 4 and given zeros, we can use a factored form of a polynomial. The factored form of the polynomial can be written as:
P(x) = (x - r1)(x - r2)(x - r3)(x - r4)
Where r1, r2, r3, and r4 are the zeros of the polynomial. In this case, the given zeros are 4, -3i, 1, and 1.
However, we need to consider that the given zero, -3i, is a complex number. Complex zeros always come in pairs, which means -3i has a corresponding zero of 3i. Hence, the zeros become: 4, -3i, 3i, 1, and 1.
Now we can proceed to form the polynomial using these zeros and considering the given multiplicity of 2 for the zero of 1 (i.e., it appears twice).
P(x) = (x - 4)(x - (-3i))(x - 3i)(x - 1)(x - 1)
To simplify, we can multiply the complex conjugate pairs together:
P(x) = (x - 4)(x + 3i)(x - 3i)(x - 1)(x - 1)
Now we can expand the equation to get the standard form of the polynomial:
P(x) = (x^2 - 4x + 3ix + 12)(x^2 - 2x + 1)
Multiplying each term, we get:
P(x) = (x^4 - 6x^3 - 5x^2 + 28x - 12)(x^2 - 2x + 1)
Finally, simplifying, we get the polynomial:
P(x) = x^6 - 8x^5 + 20x^4 - 38x^3 + 43x^2 - 40x + 12
So, the polynomial with a degree of 4, zeros 4, -3i, 1 (with multiplicity 2), and its corresponding complex conjugate, are given by P(x) = x^6 - 8x^5 + 20x^4 - 38x^3 + 43x^2 - 40x + 12.
If the polynomial has real coefficients, then the complex roots occur in conjugate pairs, so +3i is also a root. So,
your description is a bit muddy, but I'd say
y = (x-4)(x^2+9)(x-1)
That is degree 4. If you want x=1 to be of multiplicity 2, then that would be
y = (x-4)(x^2+9)(x-1)^2
Bu that is of degree 5, so maybe you don't need real coefficients, making possible
y = (x-4)(x+3i)(x-1)^2