Question

How do you factor `8n ^ { 2} - 9n - 4`?

Answer 1

`8n^2-9n-4 = 8(n-9/16-sqrt(209)/16)(n-9/16+sqrt(209)/16)`

Explanation:

Given:

`8n^2-9n-4`

Note that this is in the form:

`an^2+bn+c`

with `a=8`, `b=-9` and `c=-4`

Let us look at the discriminant `Delta` to see what we have:

`Delta = b^2-4ac`

`color(white)(Delta) = (color(blue)(-9))^2-4(color(blue)(8))(color(blue)(-4))`

`color(white)(Delta) = 81+128`

`color(white)(Delta) = 209`

`color(white)(Delta) = 11*19`

So `Delta > 0` but is not a perfect square.

As a result we can deduce that the given quadratic will factor with irrational real coefficients (but not with rational ones).

We can deduce some factors from the zeros of the quadratic, which we can find using the quadratic formula:

`n = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(n) = (-b+-sqrt(Delta))/(2a)`

`color(white)(n) = (-(color(blue)(-9))+-sqrt(Delta))/(2(color(blue)(8)))`

`color(white)(n) = 9/16+-sqrt(209)/16`

Hence, we can factor the given quadratic as:

`8n^2-9n-4 = 8(n-9/16-sqrt(209)/16)(n-9/16+sqrt(209)/16)`

Answer 2

`8n^2-9n-4=8(n-9/16+sqrt209/16)(n-9/16-sqrt209/16)`

Explanation:

Another way of factorizing is completion of square.

`8n^2-9n-4`

= `8(n^2-9/8n-4/8)`

= `8(n^2-2xxcolor(red)(9/16)xxn+color(red)((9/16)^2)-color(blue)(81/256)-4/8)`

= `8((n-9/16)^2-(81+4xx32)/256)`

= `8((n-9/16)^2-209/256)`

= `8((n-9/16)^2-(sqrt209/16)^2)`

= `8(n-9/16+sqrt209/16)(n-9/16-sqrt209/16)`