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=(-8+/-√(8^2-4(1)(17)))/(2)`, That simplifies down to
`x=(-8+/-√(64-68))/(2)`, and
`x=(-8+/-√(-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=(-8+/-√(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=(-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

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

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)]