How to use synthetic and factor theorem in application question and hard homework questions.

Hi…..today we didn’t learn much so I’m just going to go over the homework we took up in class today and the word problem we had at the end of the class

 

There were 2 main questions that people didn’t get, number 9 and 18.

 Number 9:

Question: x^3 -2kx^2 + 6x -4 is divisible by (x+ 2) with a remainder of 0. Find the value of k

 

We discussed two ways to figure out this problem.

 

First method:

 

            (x+2) = 0

            x = -2

 

so   f(-2) = 0    for the equation of   x^3 -2kx^2 + 6x -4

 

then we plug in (-2) in for all the “x”s in the equation

 

            0 = (-2)^3 - 2k(-2)^2 + 6(-2) -4

 

Solve for k

 

            k = (-8-12-4) /8

           

            k =    -3

 

and that’s how you do it for this method

This second method was thought up by Sharon  (I think).

 

Second method:

 

Her way is basically doing synthetic backwards. Since we know that the remainder is going to be 0 then the last number should be 0. So we set up the synthetic like we always do, but this time we know the remainder ahead of time.

 

 

-2 |     1  -2k  +6  -4

+

   ------------------------         

                             0R

                         

 

In order for the remainder to be 0 some number plus -4 has to equal 0. So obviously that number is +4.

 

So:

 

-2 |     1  -2k  +6  -4

 +                           4           

   --------------------------             

                             0R

         

                   

So now we divide 4 by -2 (since we are doing it backwards), which is -2. So something plus +6 has to equal 2 which is -8

 

 

-2 |     1  -2k  +6  -4

 +                    -8 +4           

     ------------------------              

                      -2   0R

 

So now we do the same thing as above and take (-8) from the second row and divide it by (-2) which gives us +4 so now we need to add something to -2k to make it +4. But we can’t know what is “k’ yet, so how can we possibly find something that would add up with -2k to equal +4. So we look more to the left and we realize that we can just bring the 1 down.

 

-2 |     1  -2k  +6  -4

 +                    -8 +4           

 ----------------------------- 

          1    +4   -2   0R

 

Since we had the one on the bottom we can use it to do synthetic the right way again (not backwards). So we bring the 1 down and multiply it by 2 so we get 2, which we place in the empty under the -2k

 

 

-2 |     1  -2k  +6  -4

 +            -  2  -8 +4           

--------------------------------                   

          1    +4   -2   0R

 

 Now that we completed the synthetic with a unknown value (-2k) we can use this to find the unknown. We know that -2k - 2 = 4

 

-2k - 2 = 4

 

k= (4+2)(-0.5)

 

k= -3

 

And that is the second method=)

 

 

 

Number 18

 

Question:  f(x) = 2x^3 + mx^2 + nx -3   and   f(x) = x^3 -3mx^2 + 2n +4    both can be divided by (x-2) with a remainder of 0 

 

 

First we sub in +2 for all the “x”s.

 

f(x) = 2x^3 + mx^2 + nx -3

 

f(2) = 16 + 4m + 2n -3

 

0 = 4m + 2n + 13

 

and

 

 

f(x) = x^3 -3mx^2 + 2n +4

 

f(2)= 8 – 12m + 4n + 4

 

0= 12m - 4n -12

 

 

We have two unknowns and two equation so now we can use substitution or elimination or whatever thing you want to use. I like using elimination because in substitution it is easier to make mistakes. ( I think, I don’t know about you). So I’ll use elimination by first setting it up.

 

12m – 4n -12 = 0

4m  + 2n + 13 = 0

 I then multiply the second equation (bottom one)  by 2.

 

    12m – 4n -12 = 0

+ 8m  + 4n + 26 = 0(this one was mutiplied by 2)

   20m + 0n  14  = 0

 

 

Now that we have only one unknown we can solve for it.

 

20m + 14 = 0

 

m = -14/20   

 

Then we sub “m” back into one of the equation to find “n”

 

 

12(-14/20) -4n -12 = 0

 

-4n = 12 - 12(-14/20)

n = -[12 – 12(-14/20)]/4

n = -5.1

 

Therefore the value of m is -0.7 or -14/20 and the value of n is -5.1

 

 

 

The question from today’s lesson that we did in class:

It said that the volume in cm^3 of ice can be represented by the function of

 V(x) = 9x^3 + 60^2 + 249x, where x represents the thickness of the ice block in cm.

It wants you to find the maximum thickness of this ice sculpture can have with a volume of 2532 cm^3.

 

2532cm^3 I the volume is represented by this V(x). So 2532cm^3 = V(x).

So:

 

2532 = 9x^3 + 60^2 + 249x

 

 9x^3 + 60^2 + 249x -2532 = 0

 

Now you just use the factor theorem to find a root of the function. We found in class that (x-4) was a root. So now we do synthetic dividing  9x^3 + 60^2 + 249x -2532 by (x-4)

 

 

4|    9   60   249   -2532

+          36   384     2532

   -----------------------------    

        9   96   633     0R

 

 

so the factored form of  9x^3 + 60^2 + 249x -2532  = 0 would be:

 

= (x – 4)(9x^2+96x+633)

 

= 3(x – 4) (3x^2+32x+211)

 

 

from the we can tell that (3x^2+32x+211) is not factorable.

 

x = [-32 +/- [(32^2 – (4)(3)(211)]^0.5]/ (2)(3)

 

x = [-32 +/- [(1024 – 2532]^0.5]/ 6

 

x = [-32 +/- (-1508)^0.5]/ 6

 

right now for our level of math we cannot take the root of a negative number. In this case -1508. So there is only one real root of 9x^3 + 60^2 + 249x -2532  = 0 and that root is 4 so the maximum thickness of the block when the volume is 2532cm^3 is 4inches.

 

THE END