<em>6 + 7r - 4 </em>× <em>11
</em>=<em> 7r + (6-4) </em>× <em>11
= 14r+11</em>
I don't think this is possible...
But if you mean 9 then the root is 3
Answer:
(x+3)(x+4)(x-4)
x+2 and x-3 are not factors but x-4 is
Step-by-step explanation:
let's factor it :)
x^3 + 3x^2 - 16x -48
first we will factor this:
x^3 + 3x^2
x^2(x + 3)
then factor the second part :
- 16x - 48
-16(x + 3)
so now,
x^2(x + 3) - 16(x + 3)
factor out x+3
(x+3)(x^2 - 16)
factor x^2 - 16
(x+3)(x+4)(x-4)