How do you find the zeroes of f(x)=6x4−5x3−12x2+5x+6?
1 Answer
Jun 24, 2016
Explanation:
f(x)=6x4−5x3−12x2+5x+6
First note that the sum of the coefficients is
That is:
6−5−12+5+6=0
Hence
6x4−5x3−12x2+5x+6=(x−1)(6x3+x2−11x−6)
If you reverse the signs of the terms of odd degree on the remaining cubic factor, then the sum is
That is:
−6+1+11−6=0
Hence
6x3+x2−11x−6=(x+1)(6x2−5x−6)
To factor the remaining quadratic expression, use an AC method:
Find a pair of factors of
The pair
Use this pair to split the middle term and factor by grouping:
6x2−5x−6
=6x2−9x+4x−6
=(6x2−9x)+(4x−6)
=3x(2x−3)+2(2x−3)
=(3x+2)(2x−3)
Putting it all together:
6x4−5x3−12x2+5x+6=(x−1)(x+1)(3x+2)(2x−3)
