Find the remaining solutions.

f(x) = x³- 2x² + 5x - 10 ; x = 2

You would have to synthetically divide f(x) by x = 2, then use the quadratic formula to find the two remaining solutions as 

x = i √5 and x = - i √5

ALL the solutions: x = 2, x = i √5 and x = - i √5

If you found solutions to a polynomial to be 

x= 3, x = - 2, and x =1

What would be the factored form equation look like?

f(x) = (x -3)(x+2)(x-1)

You are given a polynomial

f(x) = x³ - 5x² + 3x - 13. 

You are also given (x - 4)

Is that a factor or a solution?

Factor, because it has parenthesis around it. (*** Just be careful because NOT all factors have parenthesis around it) 

The solution would be x = 4.

What if you are asked to find ALL the solutions to a polynomial but you are only given 

f(x) = x³ - 2x² - 11x + 2. 

You can graph the polynomial and locate a rational zero(s)

(in other words a lattice point). 

Then you would divide the polynomial by the rational zero found from the calcualtor through synthetic division. 

Lastly, quadtratic formula can be used to find the last two solutions. 

How would you find the solution to the following equations?

f(x) = x³ + x² - 4x - 4 and f(x) = -1

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You would enter one equation into y1 and one equation into y2 and use 2nd, trace, intercept to find their intersection point(s). 

and then you would graph the two equations. ( Example 3 from 6.3AB notes)