How do you factor 5x2−3x+4?

1 Answer
George C.
Mar 3, 2018

5x2−3x+4=120(10x−3−√71i)(10x−3+√71i)

Explanation:

Given:

5x2−3x+4

We can factor with non-real complex coefficients by completing the square and using the difference of squares identity:

A2−B2=(A−B)(A+B)

with A=10x−3 and B=√71i

I will first multiply by 22⋅5=20, then divide by 20 at the end - in order to avoid much arithmetic with fractions.

20(5x2−3x+4)=100x2−60x+80

20(5x2−3x+4)=(10x)2−2(10x)(3)+32+71

20(5x2−3x+4)=(10x−3)2+(√71)2

20(5x2−3x+4)=(10x−3)2−(√71i)2

20(5x2−3x+4)=((10x−3)−√71i)((10x−3)+√71i)

20(5x2−3x+4)=(10x−3−√71i)(10x−3+√71i)

So:

5x2−3x+4=120(10x−3−√71i)(10x−3+√71i)

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