A polynomial f(x) with real coefficients and leading coefficient 1 has zeros 6 + 4i, -5 + i and degree 4. Express f(x) as a product of quadratic polynomials with real coefficients that are irreducible over R.
ive done this problem every which way i could. specifically i understand that imaginary numbers come in pairs so i wrote: 6+4i, 6-4i, -5+i, -5-i and i put all thos equal to x, subtracted the x so the equation is equal to zero and then began to multipy them .
can someone please write out the steps for me and then give me the format this answer is supposed to be in..PLEASE HELP!
(x-6+4i)(x-6-4i)(x+5+i)(x+5-i) = 0
(x^2 -12 x + 52)(x^2 + 25 x + 26)
x^4 +13 x^3 -222 x^2 + 988 x +1352
To express the polynomial f(x) as a product of quadratic polynomials with real coefficients, we know that each complex zero must be paired with its conjugate. In this case, the given complex zeros are 6 + 4i, 6 - 4i, -5 + i, and -5 - i.
Step 1: Pair the complex zeros:
Pairing the complex zeros gives us the following pairs:
(6 + 4i) and (6 - 4i)
(-5 + i) and (-5 - i)
Step 2: Set up the quadratic polynomials:
For each pair, we can set up a quadratic polynomial by using the conjugate as the second factor. Let's call the factors A and B:
For the pair (6 + 4i) and (6 - 4i):
A = (x - (6 + 4i))
B = (x - (6 - 4i))
For the pair (-5 + i) and (-5 - i):
A = (x - (-5 + i))
B = (x - (-5 - i))
Step 3: Simplify the quadratic polynomials:
Expanding the factors A and B, we get:
For the pair (6 + 4i) and (6 - 4i):
A = (x - 6 - 4i)
B = (x - 6 + 4i)
For the pair (-5 + i) and (-5 - i):
A = (x + 5 - i)
B = (x + 5 + i)
Step 4: Multiply the quadratic polynomials:
Multiply the paired quadratic polynomials together:
(A)(B) = (x - 6 - 4i)(x - 6 + 4i) = x^2 - 12x + 36 + 16
(A)(B) = (x + 5 - i)(x + 5 + i) = x^2 + 10x + 25 + 1
Step 5: Combine the quadratic polynomials:
Combine the multiplied quadratic polynomials:
f(x) = (x^2 - 12x + 52)(x^2 + 10x + 26)
Therefore, the expression for f(x) as a product of quadratic polynomials with real coefficients that are irreducible over R is:
f(x) = (x^2 - 12x + 52)(x^2 + 10x + 26)
To express the polynomial f(x) as a product of quadratic polynomials with real coefficients, we need to consider that complex (imaginary) roots always come in conjugate pairs. Since the given zeros are 6 + 4i, 6 - 4i, -5 + i, and -5 - i, we can pair them in the following way:
(6 + 4i) and (6 - 4i)
(-5 + i) and (-5 - i)
To find the quadratic polynomials, we use the fact that for a given complex zero a + bi, where a and b are real numbers, the corresponding quadratic polynomial has factors (x - (a + bi)) and (x - (a - bi)).
So, for the pair (6 + 4i) and (6 - 4i), the corresponding quadratic polynomial is:
[(x - 6 - 4i)(x - 6 + 4i)]
Expanding this out, we get:
[(x - 6)^2 - (4i)^2]
Simplifying further:
[(x^2 - 12x + 36) - (16)]
Combining like terms:
[x^2 - 12x + 20]
Similarly, for the pair (-5 + i) and (-5 - i), the corresponding quadratic polynomial is:
[(x + 5 - i)(x + 5 + i)]
Expanding this out and simplifying gives:
[x^2 + 10x + 26]
Now, we can express f(x) as a product of the two quadratic polynomials:
f(x) = (x^2 - 12x + 20)(x^2 + 10x + 26)
To finalize the form of the answer, we can multiply the factors out. This yields a fourth-degree polynomial in the standard form with real coefficients as the final answer.
Therefore, the expression for f(x) as a product of irreducible quadratic polynomials with real coefficients is:
f(x) = x^4 - 2x^3 - 12x^2 + 136x + 520.