Expert: Scott A Wilson - 7/22/2010

Question
√= root

1) (x^2 - x + 3)/(x+1)

9) (x^6-1)/(x+1)

21)2x^4 - 5x^3 - 11x^2 + 20x + 12; -2, 3
a){-0.5, 2}
b){±2, 3, -0.5}
c){0.5, 2}
d){0, 3}

22)Find an equation with integral coefficients that has the given roots:1, -3, -4
a)x3 + 12x2 + 5x - 12 = 0
b)x3 - 6x2 + 5x - 12 = 0
c)x3 + 6x2 + 5x - 12 = 0
d)x3 - 12x2 + 10x - 24 = 0

23)Determine the missing roots:

Equation:  x^4 - 3x^3 + 4x^2 - 6x + 4 = 0

Roots:   

1, 2, i√2
a)-i√2
b)-i
c)-i√5
d)i√2  

25) Solve for x
(x + 1 /-x+2) = 2/x
a)x = ±4
b)x = 1,-4
c)x = -1, 3
d)x = 1, 4

31) (4u^4 - 4u^3 - 5u^2 - 9u - 1)/ (2u - 1)   

33) (2x^4 - x^3 + x^2 + 4x - 4)/ (2x^2 + x - 2)

Answer
Since the answer you get might be these answers, I bet we got the same ones.
Still, check them to make sure they are correct.

1) (x^2 - x + 3)/(x+1)
Doing long division with the coefficient of x² in the hundreds spot, the coefficient of x in the ten's spot, and 3 in the one's spot gives us (1 1) divided into (1 -1 3).
The number we are dividing has a leading 1, so the first part of the answer is 1.

Subtracting one times (1 1) from the front of (1 -1 ...) gives us (0 -2...), so now there is, dropping the leading 0, a (-2 3) being divided by (1 1).  From here, multiply (1 1) by -2 and get (-2 -2) and subtract this from (-2 3) giving (0 5).

So the answer would by (x+1)(x-2) + 5.


9) (x^6-1)/(x+1)
This can be done in a similar way.
We are dividing (1 1) into (1 0 0 0 0 -1).
The first term in the answer is 1 since subtracting (1 1) from (1 0...) gives us (0 -1...).
The number we are dividing into now becomes (-1 0 0 0 -1) since the leading 0 is dropped.
Taking (1 1) into (-1 0 ...) gives us a multiplier of -1.  Now -1(1 1) = (-1 -1), and (-1 0)
minus (-1 -1) gives us (0 1 ...), so again we drop the leading 0 and get (1 0 0 1).  This can be repeated until we get (1 -1 +1 -1 +1 -1) which means we have (x+1)(x^5 - x^4 + x^3 - x^2 + x -1).


21)2x^4 - 5x^3 - 11x^2 + 20x + 12; -2, 3

I assume we are looking for the other two solutions?

Since -2 is a solution, we can divide by (x+2). This gives (x+2)(2x^3 - 9x^2 + 7x +6).
Since 3 is a solution, we can divide by (x-3).  This gives (x+2)(x-3)(2x^2-3x-2).
The last term is 2x^2 - 3x - 2, so a=2, b=-3, and c=-2; the solution is the quadratic equations.
That is, (-b ±√(b²-4ac))/(2a).  That becomes (3±√(9+16))/4, which is (3±√25)/4.
It is known that √25=5, so the answer is (3±5)/4, and that is -2/4 and 8/4.
The first number reduces to -1/2, which is -0.5, and the second number reduces to 2.

a){-0.5, 2}
b){±2, 3, -0.5}
c){0.5, 2}
d){0, 3}


22)Find an equation with integral coefficients that has the given roots:1, -3, -4
This means the function is (x-1)(x+3)(x+4).  Now (x+3)(x+4) is x² + 7x + 12.
Multiplying that by (x-1) gives x³ + 6x² + 5x - 12.  You see it too?

a)x3 + 12x2 + 5x - 12 = 0
b)x3 - 6x2 + 5x - 12 = 0
c)x3 + 6x2 + 5x - 12 = 0
d)x3 - 12x2 + 10x - 24 = 0


23)Determine the missing roots:

Equation:  x^4 - 3x^3 + 4x^2 - 6x + 4 = 0

Roots: 1, 2, i√2
Divide by (x-1) and (x-2).
Dividing by (x-1) gives (x³-2x²+2x-4), so it is (x-1) (x³-2x²+2x-4).
Dividing (x³-2x²+2x-4) by (x-2) gives us (x²-2) { note that there is no x term }.
This makes the other two solutions into ±i√2.
Since we are given one of the solutions is i√2, the other is -i√2.  See that answer?

a)-i√2
b)-i
c)-i√5
d)i√2  


25) Solve for x: (x + 1 /-x+2) = 2/x
Do you mean (x+1)/(-x+2) = 2/x?
Multiply through on both sides by x(-x+2).
This gives x² + x = - 2x + 4.
Subtracting off the left side { -2x+4 } from both sides gives x² + 3x - 4 = 0.
This facors into (x-4)(x+1)= 0, so the solutions are x=4 and x=-1.  Yeah, that's a choice.

a)x = ±4
b)x = 1,-4
c)x = -1, 3
d)x = 1, 4


31) (4u^4 - 4u^3 - 5u^2 - 9u - 1)/ (2u - 1)   
Again, synthetically, divide (2 -1) into (4 -4 -5 -9 -1).
This is done by dividing the first integer of the long expression by 2, and that gives the multiplier.  The multiplier is the first number in the answer, and subtracting the result
gives a 0 up front and changes the next term, a -4, into a -4 - (-2)(-1) = -6.

Next, take (-6 -5 -9 -1) and divide that be (2 -1).

33) (2x^4 - x^3 + x^2 + 4x - 4)/ (2x^2 + x - 2)
Here, I could show a different approach.
Take (2x² + x - 2)(ax² + bx + c) = 2x^4 - x^3 + x^2 + 4x - 4.
We need to solve the equations for x^4, x^4, x^2, x, and the constant.
For x^4, the equation is 2a = 2, so a=1.

This gives us  (2x² + x - 2)(x² + bx + c) = 2x^4 - x^3 + x^2 + 4x - 4.
The next term would be given by 2b + 1 = -3, since these are the coefficients for x³.
This says that 2b = -2, or b = -1.
Putting this in gives (2x² + x - 2)(x² - x + c) = 2x^4 - x^3 + x^2 + 4x - 4.

For x², there wold be three terms.  They are 2c - 1 - 2.
Since the coefficient on x² is 1, we have 2c - 1 - 2 = 1.
That is the same as 2c - 3 = 1, or 2c = 4, or c = 2.
This said, then lets try (2x² + x - 2)(x² - x + 2) = 2x^4 - x^3 + x^2 + 4x - 4.
Multiply it out and the answer should be gotten.