Question

How do you factor `16x ^ { 2} - 40x + 2`?

Answer 1

`(x - (5 + sqrt 23)/4)(x - (5 - sqrt 23)/4) * 16`

Explanation:

`8x^2 - 20x + 1 = 0`

`Delta = 400 - 4 * 8 = 368 = 2^4 * 23`

`x = (20 ± 4 sqrt 23)/16 = (5 ± sqrt 23)/4`

Answer 2

`f(x) = 16(x - 5/4 - sqrt23/4)(x - 5/4 + sqrt23/4)`

Explanation:

`f(x) = 16x^2 - 40 x + 2 = 2y = 2(8x^2 - 20x + 1)`
Factor `y = 8x^2 - 20 x + 1`
`D = d^2 = b^2 - 4ac = 400 - 32 = 368` --> `d = +- 4sqrt23`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = 20/16 +- (4sqrt23)/16`
`x1 = 5/4 + sqrt23/4`, and
`x2 = 5/4 - sqrt23/4`
The factored form of y is:
`y = a(x - x1)(x - x2) =`
`y = 8(x - - 5/4 - sqrt23/4)(x - 5/4 + sqrt23/4)`
Finally,
`f(x) = 16(x - 5/4 - sqrt23/4)(x - 5/4 + sqrt23/4)`

Note. You may apply the expression of the quadratic function in intercept form (Google Search):
`y = a(x - b/(2a) - d/(2a))(x - b/(2a) + d/(2a))`