Question

Find the complex factors of the quadratic trinomial x^2+8x+17

Answer 1

You will plug these numbers into the quadratic formula to solve. a = 1, b = 8, c = 17. $x=\frac{-8+/-\sqrt{8^2-4(1)(17)}}{2}$, That simplifies down to $x=\frac{-8+/-\sqrt{64-68}}{2}$, and $x=\frac{-8+/-\sqrt{-4}}{2}$. We say our roots are complex because we have a negative under the square root sign which we know is illegal. Unless we use the imaginary "i" and offset that negative. -1 is equal to i-squared, so let's make that replacement. $x=\frac{-8+/-\sqrt{4*i^2}}{2}$. Both 4 and i-squared are perfect squares that can be pulled out as a 2 and an i, respectively. $x=\frac{-8+/-2i}{2}$. Everything in the numerator can be reduced by the 2 in the denominator to give us these 2 solutions for x: $x=-4+i$ and $x=-4-i$.

Answer 2

Answer:

[x+(4-i)][x+(4+i)]

Step-by-step explanation:

Complete the square then regroup the first two terms. Add and subtract: (b/2)^2=(8/2)^2=16

x^2+8x+17=(x^2+8x)+17

=(x^2+8x+16)+17-16

=(x+4)^2+1  create a difference of squares using i^2= -1

=(x+4)^2 - (-1)

=(x+4)^2 - i^2  use the difference of two squares identity

=[(x+4) - i][(x+4)+i]

=[x+(4-i)][x+(4+i)]