How do you factor completely: #35x^2-54x+16#?

2 Answers
Aug 8, 2015

Answer:

Use quadratic formula to find roots of #35x^2-54x+16 = 0# and deduce:

#35x^2-54x+16 = (5x-2)(7x-8)#

Explanation:

Let #f(x) = 35x^2-54x+16#

This is of the form #ax^2+bx+c#, with #a=35#, #b=-54# and #c=16#.

#f(x) = 0# has roots given by the quadratic formula:

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

#=(54+-sqrt((-54)^2-(4xx35xx16)))/(2xx35)#

#=(54+-sqrt(2916 - 2240))/70#

#=(54+-sqrt(676))/70#

#=(54+-26)/70#

That is #x = 28/70 = 2/5# or #x = 80/70 = 8/7#

So #(5x-2) = 0# or #(7x-8) = 0#

Hence:

#35x^2-54x+16 = (5x-2)(7x-8)#

Answer:

Factor #y = 35x^2 - 54x + 16#

Ans: #(5x - 2)(7x - 8)#

Explanation:

I use the new AC Method to factor trinomials (Google, Yahoo Search)

#y = 35x^2 - 54x + 16 = #35(x + p)(x + q)

Converted trinomial: #y' = x^2 - 54x + 560.#
Find 2 numbers p' and q' knowing sum (-54 = b) and product (ac = 560)'. Here p' and q' have same sign.
Factor pairs of ac = 560 --> (10, 56)(14, 40). This sum is 54 = -b. Change the sum to the opposite.
Then p' = -14, and q' = -40, Therefore,
#p = (p')/a = -14/35 = -2/5# and #q = (q')/a = -40/35 = -8/7#.

Factored form:

#y = 35(x - 2/5)(x - 8/7) = (5x - 2)(7x - 8)#