Question

Hello I need help with a problem ,8x^2-9x-3It says factor completely

Answer 1

Not factorable

Step-by-step explanation

given

`8x^2-9x-3`

Step 1

given

`8x^2-9x-3`

use the quadratic formuña

it says

`\begin{gathered} for \\ ax^2+bx+c=0 \\ the\text{ solution for x is} \\ x=(-b\pm√(b^2-ac))/(2a) \end{gathered}`

so

If Δ=0 then ax2+bx+c is a perfect square trinomial, expressible as

(√ax+√c)2 or as (√ax−√c)2 .

If Δ<0 then ax2+bx+c has two distinct Complex zeros and is not factorable over the reals. It is factorable if you allow Complex coefficients

`\begin{gathered} \Delta=b^2-4ac \\ \Delta=(-9)^2-4(8)(-3) \\ \Delta=81+96 \\ \Delta=177 \end{gathered}`

then, as

`\Delta>0`

the expression has two distinct Real zeros and is factorable over the Reals

so,

the expression has real solution ,so

Not factorable

I hope this helps you