2x3 - 11x2 - 16x + 105 = 0

Question

Find all solutions of the equation.

Answer 1

2x³-11x²-16x+105=0

possible solutions are 2 and the factors of 105 which are 3, 5, and 7 (and their opposites)

using trial and error or synthetic division(the better choice), you find that 5 is a solution

 

(2x³-11x²-16x+105)/(x-5)=2x²-x-21

2x²-x-21=(2x-7)(x+3)

(x-5)(2x-7)(x+3)=0

x=5, x=7/2, and x=-3

Answer 2

Hey Colsten,

 

2x³ - 11x² - 16x + 105 = 0

 

There will be 3 real zeros since it is 3rd power.  This equation will be factored like this:

 

As you see +105, you can conclude that constants are sharing same sign.  To find out the signs for constants, look at the middle term.  In this case, middle terms are negative.  So, 2 multiples to form +105 are negative.  So, what are they? (1, 105) (3, 35) (5, 21) (7, 15)

 

I picked (5, 21) because of -16x. 

 

(2x² - x - 21) (x - 5) = (2x - 7) (x + 3) (x - 5) = 0

 

If I put 21 into with binomial (x - 21), then my middle term will be in the 40s.  Therefore, I had to put 21 inside of quadratics with 2x².  After that, I get -16x but what about -11x²? since (2x²)* (-5) is -10x², all I need is -x².  To achieve that, all I needed to do was include -x with 2x² and -21: (2x² - x - 21).  And other one is (x - 5).

 

Since (2x² - x - 21) is factorable, it needs to be factored.  Factors for 21 will be 7 & 3.  As you can see right away, 2 & 3 makes 6 and 7 & 1 makes 7.  And middle term is -x.  Since 3 has to be multiplied, it must be outside and 7 will be joined with 2x.  So, it will look like (2x - 7) (x + 3) for (2x² - x - 21).

 

In conclusion, it will looked like this in factored form:

 

2x³ - 11x² - 16x + 105 = 0

(2x² - x - 21) (x - 5) = (2x - 7) (x + 3) (x - 5) = 0

 

x = +7/2, -3, +5