• Oct 12th 2007, 09:01 AM
cathystntn123
Can anyone help me with these three sums!!

(x+5)(x-4)(x-a)=xto the cube+bx to be square-23x+c=0
Find the values of a,b, and c

x to be squard+6x-k is afactor of x to be cube+kx to be squared-x-k!!Find the value of k!!Hence solve the following equation x to cubed+kx to be squared-x-k=0

(x to be squared+ax+b) is a factor of 2x to cube+ax to be squared+13x+b

Find the values of a and b

Thanks alot!!
• Oct 12th 2007, 09:38 AM
DivideBy0
Here are my ideas for the questions...

In question 1, expanding we get

$\displaystyle (x^2+x-20)(x-a)=x^3+bx^2-23x+c=0$

$\displaystyle x^3-ax^2+x^2-ax-20x+20a=x^3+bx^2-23x+c=0$

$\displaystyle x^3+(1-a)x^2-(a+20)x+20a=x^3+bx^2-23x+c=0$

So we have

$\displaystyle 1-a=b$ ...[1]

$\displaystyle a+20=23$ ...[2]

$\displaystyle 20a=c$ ...[3]

From [2],
$\displaystyle a=3$
Then from [1]
$\displaystyle b=-2$
And from [2]
$\displaystyle c=60$

For question 2, $\displaystyle x^2+6x-k | x^3+kx^2-x-k$

If you factor $\displaystyle x^2+6x-k$ you get
$\displaystyle x^2+6x-9+9-k=(x+3)^2-(9+k)=(x+3+\sqrt{9+k})(x+3-\sqrt{9+k})$
Then by the factor theorem...

$\displaystyle (-3-\sqrt{9+k})^3+k(-3-\sqrt{9+k})^2-(-3-\sqrt{9+k})-k=0$

and

$\displaystyle (-3+\sqrt{9+k})^3+k(-3+\sqrt{9+k})^2-(-3+\sqrt{9+k})-k=0$

You should be able to solve either for k...

Haha! :D My calculator got k = 7 after a very... long time. And I had to resort to iteration. Surely there is a more efficient way to do this.
• Oct 12th 2007, 10:28 AM
topsquark
Quote:

Originally Posted by cathystntn123
x to be squard+6x-k is afactor of x to be cube+kx to be squared-x-k!!Find the value of k!!Hence solve the following equation x to cubed+kx to be squared-x-k=0

We know that $\displaystyle x^2 + 6x - k$ is a factor of $\displaystyle x^3 + kx^2 - x - k$.

Since the quadratic is a factor of the cubic we must have that
$\displaystyle (x^2 + 6x - k)(x - a) = x^3 + kx^2 - x - k$

$\displaystyle x^3 + (6 - a)x^2 - (6a + k)x + ak = x^3 + kx^2 - x - k$

Matching coefficients we know that
$\displaystyle 6 - a = k$
and
$\displaystyle -6a - k = -1$
and
$\displaystyle ak = -k$

Rewriting the first equation gives:
$\displaystyle k = 6 - a$

So
$\displaystyle -6a - (6 - a) = -1$
and
$\displaystyle a(6 - a) = -(6 - a)$

The first of these says:
$\displaystyle -6a - 6 + a = -1 \implies -5a = 5 \implies a = -1$

The second of these says:
$\displaystyle 6a - a^2 = 6a + a \implies a^2 + a = 0 \implies a = -1, 0$

The only solution common to both of these is that a = -1. Thus
$\displaystyle k = 6 - (-1) = 7$.

Thus
$\displaystyle (x^2 + 6x - 7)(x + 1) = x^3 + 7x^2 - x - 7$
which you can verify.

So now we solve
$\displaystyle x^3 + 7x^2 - x - 7 = 0$

$\displaystyle (x + 1)(x^2 + 6x - 7) = 0$

$\displaystyle (x + 1)(x + 7)(x - 1) = 0$

So $\displaystyle x = -7, -1, 1$

-Dan
• Oct 12th 2007, 10:37 AM
topsquark
Quote:

Originally Posted by cathystntn123
(x to be squared+ax+b) is a factor of 2x to cube+ax to be squared+13x+b

Find the values of a and b

Use the same process as I used before:
$\displaystyle (x^2 + ax + b)(2x + c) = 2x^3 + ax^2 + 13x + b$
(Why do we need the "2" on the linear factor? Because we know the coefficient of the leading term of the cubic is a 2 and this is the only place this factor can come from.)

This will give you three equations in three unknowns. What I would recommend is that you solve for c first, then put b in terms of a. The result will be a cubic equation for a that factors:
$\displaystyle a^3 + a^2 + 13a + 13 = (a + 1)(a^2 + 13) = 0$

So I get that a = -1, b = 7, and c = 1.

(Make sure you can fill in the intermediate steps.)

-Dan
• Oct 12th 2007, 10:37 AM
red_dog
Quote:

Originally Posted by cathystntn123
x to be squard+6x-k is afactor of x to be cube+kx to be squared-x-k!!Find the value of k!!Hence solve the following equation x to cubed+kx to be squared-x-k=0

We have $\displaystyle x^3+kx-x-k=(x^2+6x-k)(x+l)$
Expand the right side: $\displaystyle x^3+kx-x-k=x^3+(l+6)x^2+(6l-k)x-kl$
Then $\displaystyle \left\{\begin{array}{lll}l+6=k\\6l-k=-1\\-kl=-k\end{array}\right.$

From the first two equations we get $\displaystyle k=7, \ l=1$ wich also satisfy the third equation.

Now, we have to solve the equation $\displaystyle x^3+7x^2=x-7=0$
$\displaystyle x^2(x+7)-(x+7)=0\Leftrightarrow (x+7)(x^2-1)=0\Rightarrow x_1=-7,x_2=-1,x_3=1$
• Oct 12th 2007, 03:09 PM
Soroban
Hello, Cathy!

I approached them differently . . . Here's #3.

Quote:

$\displaystyle 3)\;\;x^2 +ax+b$ .is a factor of .$\displaystyle 2x^3+ax^2+13x+b$
Find the values of $\displaystyle a$ and $\displaystyle b$

I used long division . . .
Code:

                                  2x  -  a                   ---------------------------       x² + ax + b ) 2x³ +  ax² + 13x  +  b                     2x³ + 2ax² + 2bx                   -----------------------                           -ax² + (13-2b)x + b                           -ax²  -  a²x  -  ab                           --------------------                               (13-2b+a²)x + (b+ab)

Since the division "comes out even", the remainder is zero.

Hence: .$\displaystyle \begin{array}{cccccccc}13-2b+a^2& = & 0& \Rightarrow & a^2+13 & = & 2b & {\color{blue}[1]}\\b+ab & = & 0 & \Rightarrow & b(a+1) & = & 0 & {\color{blue}[2]}\end{array}$

From [2], we have: .$\displaystyle b = 0$ .or .$\displaystyle a = -1$

If $\displaystyle b=0$, [1] becomes: .$\displaystyle a^2+13\:=\:0$ . . .which has no real roots

If $\displaystyle a=-1$, [1] becomes: .$\displaystyle 1+13 = 2b\quad\Rightarrow\quad b = 7$

The problem is: .$\displaystyle (2x^3 - x^2 + 13x + 7) \div (x^2-x+7) \:=\:2x + 1$