Question

How do you factor the trinomial `r^2-34r+2`?

Answer 1

`r^2-34r+2=(r-17-sqrt(287))(r-17+sqrt(287))`

Explanation:

Factor by completing the square and using the difference of squares identity:

`a^2-b^2=(a-b)(a+b)`

with `a=(r-17)` and `b=sqrt(287)` as follows:

`r^2-34r+2`

`= (r-17)^2-17^2+2`

`= (r-17)^2-287`

`= (r-17)^2-(sqrt(287))^2`

`=((r-17)-sqrt(287))((r-17)+sqrt(287))`

`=(r-17-sqrt(287))(r-17+sqrt(287))`

Answer 2

There is another way to factor `y = x^2 - 34x + 2.`
`y = (x - 17 + sqrt287)(x - 17 - sqrt287)`

Explanation:

Use the expression of a quadratic function in intercept form
y = a(x - x1)(x - x2)
a = 1, x1 and x2 are the 2 real roots of the equation y = 0.
Solve this equation by the improved quadratic formula (Socratic Search))
`D = d^2 = b^2 - 4ac = 1156 - 8 = 1148` --> `d = +- 2sqrt287`
`x1 = -b/(2a) +- d/(2a) = 34/2 + (2sqrt287)/2 = 17 + sqrt287`
`x2 = 17 - sqrt287`
Therefor, the factored form of y is:
`y = (x - 17 - sqrt287)(x - 17 + sqrt287)`