How do you factor the expression #4s^2-22s+10#?

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Answer 1 · 2017-01-19T22:14:15

#4(x-5)(x-1/2)#

We need to find the roots of the equation:

#4s^2 - 22s + 10 = 0#.

We can first divide the entire expression by #2#:

#2s^2 - 11s + 5 = 0#. Although not necessary, this does simplify the procedure just a little bit.
Using the quadratic formula:

#x = (-b +- sqrt(b^2-4ac))/(2a)#, where

#a# is the coefficient of #x^2#
#b# is the coefficient of #x#, and
#c# is the constant term.

So, #x = (11 +- sqrt81)/4 => x = 5# or #x = 1/2#.

The formula for factoring a quadratic is:

#a(x - r_1)(x - r_2)# where #r_1,r_2# the two roots.

We still have to remember that although the roots of

#4s^2 - 22s + 10 = 0#

are the same as those of

#2s^2 - 11s + 5 = 0#,

the original value for #a# is #4#, not #2#.

So,

#4s^2 - 22s + 10 = 4(x-5)(x-1/2)#