expressing f(x) as a product of linear and quadratic polynomials when given zeros?

Question

A polynomial f(x) with real coefficients and leading coefficient 1 has the given zeros and degree. Express f(x) as a product of linear and quadratic polynomials with real coefficients that are irreducible over R.

seriously i have been trying to work this problem out for like an hour. can someone explain to me how the hell they got this answer from this problem please! any help is greatly appreciated!

zeros are 4 ±3i -2 ±i
degree is 4
answer is (x^2 -8x +25)(x^2 +4x +5)

as far as i got was:
(x+4)(x-4)(x+3i)(x-3i)(x-2)(x+2)(x+i)(x-i)
(x^2 -16)(x^2 +9)(x^2 -4)(x^2 +1)

ok, first of all did i do that right?
and where do you go from there?
i multiplyed them together but i don't see what they did to get (x^2 -8x +25)(x^2 +4x +5)

oh my wow! i read the problem wrong! i was reading it as there were 4 zeros (4 then 3i then -2 then i), but there were only 2, oops! but thank you guys for the help! i wasnt there for that day's lesson and i was left to figure it out on my own!

Answer 1

Hi,

If your zeros are:

x = 4 + 3i__x = 4 - 3i__x = -2 + i__x = -2 - i

When you move these to the left with the x, you get these four factors:

(x - 4 - 3i)(x - 4 + 3i)(x + 2 - i)(x + 2 + i)

Multiply the first 2 factors together. Then multiply the last 2 factors together. Then multiply those results together.

(x - 4 - 3i)(x - 4 + 3i) multiply together to give:

x² - 4x + 3ix -4x+ 16 - 12i - 3ix + 12i - 9i² =
x² - 8x + 16 - 9i²
Replace i² with -1 to get
x² - 8x + 16 + 9 =
x² - 8x + 25 <= This is the first factor.

(x + 2 - i)(x + 2 + i) multiply together to give:

x² + 2x + ix + 2x+ 4 + 2i - ix - 2i - i² =
x² + 4x + 4 - i²
Replace i² with -1 to get

x² + 4x + 4 - (-1) =
x² + 4x + 4 + 1 =
x² + 4x + 5 <= This is the second factor.

Now multiply these 2 factors together.
(x² - 8x + 25)(x² + 4x + 5) =
x^4 +4x³ + 5x² -8x³ - 32x² - 40x + 25x² + 100x + 125 =
x^4 - 4x³ -2x² + 60x + 125

That's it!! I hope that helps!! :-)

Answer 2

No, 4+ 3i is a COMPLETE complex root.
The roots are 4 + 3i, 4 -3i, -2+ i and -2 - i.
You cannot separate the real and imaginary parts this way!
But there is even a simpler way to do this.
Look:
4 + 3i and 4 -3i are the zeros of a quadratic factor
of your quartic.
How do you find this quadratic factor, x²+ax+b?
Well the sum of the roots of this quadratic is -a
and the product of the roots is b. (You can
get this from the quadratic formula.)
So 4+3i + 4 - 3i = 8
(4+3i)(4-3i) = 16+9 = 25.
and your quadratic factor is x²-8x+25.
Now go to work on the other quadratic factor
-2+i+-2-i = -4
(-2+i)(-2-i) = 4 + 1 = 5
So your other quadratic factor is x²+4x+5.
Finally, your quartic is the product of these 2 factors:
f(x) = (x²-8x+25)(x²+4x+5).
No tricky multiplication needed!

Answer 3

The factors are:
(x-4+3i)(x-4-3i)(x+2+i)(x+2-i)
= (x^2+4x+5)(x^2-8x+25)
What you have would give you a degree 8 equation so you know that is wrong

Answer 4

question variety one million For this polynomial equation 0*x^4 +x^3 -7*x^2 +19*x -13 = 0, answer here questions : A. sparkling up by ability of Factorization answer For question one million x^3 -7*x^2 +19*x -13 = 0 And we get P(x)=x^3 -7*x^2 +19*x -13 Now, we are able to look for the roots of P(x) utilising numerous set of rules : 1A. sparkling up by ability of Factorization x^3 -7*x^2 +19*x -13 = 0 Separate : ( x^3 -x^2 ) + ( -6*x^2 +6*x ) + ( 13*x -13 ) = 0 Commutative regulation : ( x^3 -6*x^2 +13*x ) + ( -x^2 +6*x -13 ) = 0 Distributive regulation : x*( x^2 -6*x +13 ) + -*( x^2 -6*x +13 ) = 0 factor : ( x -one million )*( x^2 -6*x +13 ) = 0 Separate : ( x -one million )*( ( x^2 +(-3+2*i)*x ) + ( (-3-2*i)*x +13 ) ) = 0 Commutative regulation : ( x -one million )*( ( x^2 +(-3-2*i)*x ) + ( (-3+2*i)*x +13 ) ) = 0 Distributive regulation : ( x -one million )*( x*( x +(-3-2*i) ) + (-3+2*i)*( x +(-3-2*i) ) ) = 0 factor : ( x -one million )*( x -3+2*i )*( x -3-2*i ) So the Polynomial have 3 roots : x1 = one million x2 = 3-2*i x3 = 3+2*i