How do you factor completely: 35x^2-54x+1635x254x+16?

2 Answers
Aug 8, 2015

Use quadratic formula to find roots of 35x^2-54x+16 = 035x254x+16=0 and deduce:

35x^2-54x+16 = (5x-2)(7x-8)35x254x+16=(5x2)(7x8)

Explanation:

Let f(x) = 35x^2-54x+16f(x)=35x254x+16

This is of the form ax^2+bx+cax2+bx+c, with a=35a=35, b=-54b=54 and c=16c=16.

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

x = (-b+-sqrt(b^2-4ac))/(2a)x=b±b24ac2a

=(54+-sqrt((-54)^2-(4xx35xx16)))/(2xx35)=54±(54)2(4×35×16)2×35

=(54+-sqrt(2916 - 2240))/70=54±2916224070

=(54+-sqrt(676))/70=54±67670

=(54+-26)/70=54±2670

That is x = 28/70 = 2/5x=2870=25 or x = 80/70 = 8/7x=8070=87

So (5x-2) = 0(5x2)=0 or (7x-8) = 0(7x8)=0

Hence:

35x^2-54x+16 = (5x-2)(7x-8)35x254x+16=(5x2)(7x8)

Factor y = 35x^2 - 54x + 16y=35x254x+16

Ans: (5x - 2)(7x - 8)(5x2)(7x8)

Explanation:

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

y = 35x^2 - 54x + 16 = y=35x254x+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)