If x² + bx -2 is factor of xcubed + (2b-1)x² - p, find two possible values of p.
Im not sure what method to use to solve this. do i divide them?
If x² + bx -2 is factor of xcubed + (2b-1)x² - p, find two possible values of p.
Im not sure what method to use to solve this. do i divide them?
Hello DetanonSuppose the other factor is $\displaystyle (rx+s)$. Then:$\displaystyle (x^2+bx-2)(rx+s) = x^3+(2b-1)x^2 -p$Comparing coefficients:
$\displaystyle x^3: r = 1$So:$\displaystyle (x^2+bx-2)(x+s) = x^3+(2b-1)x^2 -p$Comparing coefficients:$\displaystyle x^2: (s+b) = (2b-1) \Rightarrow s = b-1$ (1)So, from (1) and (2):
$\displaystyle x: -2+sb = 0 \Rightarrow sb = 2$ (2)
$\displaystyle b(b-1)=2$Constant term:
$\displaystyle \Rightarrow b^2-b-2=0$
$\displaystyle \Rightarrow (b-2)(b+1)=0$
$\displaystyle \Rightarrow b = 2, -1$
$\displaystyle \Rightarrow s = 1, -2$
$\displaystyle p = 2s$Grandad
$\displaystyle \Rightarrow p = 2, -4$
Hello DetanonBecause when one factor of a cubic is a quadratic, the other is linear; i.e. it is of the form $\displaystyle (rx + s)$ for some values of $\displaystyle r$ and $\displaystyle s$.It doesn't mean anything. You have taken it out of context. Read the whole thing. I am comparing coefficients. (Do you know what thay means?) First I am looking at the coefficient of $\displaystyle x^3$. Later on, I look at the coefficients of $\displaystyle x^2, x$ and the constant. I have used a colon (':') each time to show which coefficient I am looking at.and what does $\displaystyle
x^3: r
$mean?
No, not a method that is as sensible and straightforward as the one I have shown you.Is there a different method to do it?
Read it through again - it's really very simple.
Grandad
Hello Detanon
I really don't know what your problem is here. Multiply out the brackets from
$\displaystyle (x^2+bx-2)(rx+s)$and then note that the result has to be identically equal to
$\displaystyle x^3 +(2b-1)x^2 -p$(Can you do that?)
In this way you get the equations for $\displaystyle r, s, b$ and $\displaystyle p$ that I have shown you. I repeat: if you understand the basic rules of algebra, then trust me, this is really very simple. (And certainly much easier than trying to do an algebraic long division.)
Grandad
If $\displaystyle x^2 + bx - 2$ is a factor then it should be obvious, after a little thought, that the other factor has to be $\displaystyle x + \frac{p}{2}$. This is the only way that you can get the $\displaystyle x^3$ term and the constant term of $\displaystyle -p$.
So $\displaystyle x^3 + (2b - 1) x^2 - p = (x^2 + bx - 2)\left(x + \frac{p}{2}\right)$. Expand this out. You require the coefficient of $\displaystyle x^2$ to be $\displaystyle 2b - 1$ and the coeffcient of $\displaystyle x$ to be zero:
$\displaystyle \frac{p}{2} + b = 2b - 1 \Rightarrow p = 2b - 2$ .... (1)
$\displaystyle \frac{bp}{2} - 2 = 0 \Rightarrow bp = 4$ .... (2)
Solve equations (1) and (2) simultaneously for b and p.
You need to go back now and have a good hard think about all that's been said about this question.
For checking purposes:
Spoiler: