Question

How do you factor the trinomial `6x^2-11x-35`?

Answer 1

`color(green)((3x + 5 ) ( 2x - 7 )` is the factorised form of the expression.

Explanation:

`6x^2 - 11x - 35`

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like `ax^2 + bx + c`, we need to think of 2 numbers such that:

`N_1*N_2 = a*c = 6*(-35) = -210`

AND

`N_1 +N_2 = b = -11`

After trying out a few numbers we get `N_1 = 10` and `N_2 =-21`
`10* (-21) = -210`, and `10+(-21)= -11`

`6x^2 - 11x - 35 = 6x^2 - 21x + 10x - 35`

`= 3x ( 2x - 7 ) + 5 ( 2x - 7 )`

`(2x - 7 )` is a common factor to each of the terms

`= (3x + 5 ) ( 2x - 7 )`

`color(green)((3x + 5 ) ( 2x - 7 )` is the factorised form of the expression.

Answer 2

(3x + 5)(2x - 7)

Explanation:

Another factoring way is using the new AC Method (Socratic Search)
`y = 6x^2 - 11x - 35 =` 6(x + p)(x + q)
Converted trinomial `y' = x^2 - 11x - 210 =` (x + p')(x + q')
p' and q' have opposite signs because ac < 0.
Factor pairs of (-210) -->...(-10, 21)(10, -21). This sum is -21 = b. Then p' = 10 and q' = -21
Back to trinomial y -> p = (p')/a = 10/6 = 5/3 and q = (q')/a = -21/6 = -7/2
Factored form: y = 6(x + 5/3)(x - 7/2) = (3x + 5)(2x - 7)