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Factorice completely 80^2_45

What are the factor of 80k^2_45

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Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)

80k^2 - 45

= 5(16k^2 - 9), factoring out the GCF of the two terms, which is 5.

= 5(4k+3)(4k-3), since 16k^2 - 9 can be represented by the difference of two squares (4k)^2 - 3^2

Stuart R. | Math Tutor - Online and In Home -20 Years ExperienceMath Tutor - Online and In Home -20 Year...

Not sure i'm reading this correctly, but assuming the expression is:

80K2 - 45 then use the quadratic formula to get the roots.  (-b + sqrt(b2-4ac)/2a = formula

where a,b,c are the coefficients of the first,second,third term.

NOTE: there is NO 'b' term, just a = 80 and c = 45

so we have 0 + sqrt(0- 4(80)(-45))  all divided by 160

thus we have 0 sqrt(1440)  all divided by 160.

remember: the sqrt(1440) = sqrt(144) times sqrt(100), so we have 12 times 10

thus we have +(120)/160

Now we have TWO roots.   .75 and -.75


the factors of the expression are (k+.75) and (k-.75)

there are other ways to do this as well by just factoring, but use of the quadratic is easier to see.



I agree that the solutions are .75 and -.75 if we were trying to solve an equation; however, this is an expression and the student is being asked to factor it. Robert is definitely correct. The answer is 5(4k +3)(4k - 3), because it is completely broken down to its smallest factors using only integers.


Phillip H, NBCT