1. ## 2 Polynomial Problems

Problem 147. The numbers 1 and 2 are roots of polynomial P. Prove that P is divisible by (x-1)(x-2).

I have the solution but I don't understand one of the steps.

Solution. P is divisible by (x-1) because 1 is a root of P. Therefore $P = (x-1)\cdot{Q}$ for some polynomial Q. Substituting 2 for x in this equality we find that 2 is a root of Q, so Q is divisible by (x-2), that is $Q = (x-2)\cdot{R}$ for some polynomial R. So $P = (x-1)(x-2)\cdot{R}$

Why is it, that you can just substitute 2 into that? Am I correct in saying it shows that P = Q therefore (x-2) is a root of Q, because it's a root of P? I don't understand the reasoning behind that step. Could someone explain?

A second problem - to which I've no solution - has me stumped.

Problem 154. Assume that $x^3+ax^2+x+b$ (where a and b are some numbers) is divisible by $x^2-3x+2$. Find a and b.

2. Hello
Originally Posted by jerry
Why is it, that you can just substitute 2 into that? Am I correct in saying it shows that P = Q therefore (x-2) is a root of Q, because it's a root of P? I don't understand the reasoning behind that step. Could someone explain?
2 is a root of $P$ so $P(2)=0$
In $P(x)=(x-1)Q(x)$, we substitute $x=2$ and get $0=P(2)=1\cdot Q(2)=Q(2)$ : 2 is also a root of $Q$

Problem 154. Assume that $x^3+ax^2+x+b$ (where a and b are some numbers) is divisible by $x^2-3x+2$. Find a and b.
The first polynomial is cubic, the second quadratic so there exists $c,\,d\in \mathbb{R}$ such that $x^3+ax^2+x+b=(cx+d)(x^2-3x+2)$
Then, you can develop the RHS and identify the coefficients of the two polynomials which yields 4 equations with 4 unknowns.

3. Problem 147

A polynomial $P(x)$ has a root when $P(x)=0$, for some given value of $x$. If $1$ is a root of $P(x)$ then $P(1)=0$, hence $P(x)$ is divisible by $(x-1)$ because when $x=1$, ( $x-1)=0$. The same can be said for $x=2$.

Problem 154

If $x^3+ax^2+x+b$ is divisible by $x^2-3x+2$ then the roots of the latter must also be roots for the former for a given $x$ value.

Let $x^2-3x+2~=~0$

$(x-2)(x-1)~=~0$ therefore $x=2~,~x=1$ are roots of the equation $x^3+ax^2+x+b$

$x=1~,~~2+a+b~=~0$
$x=2~,~~10+4a+b~=~0$

Now solve simultaneously to find a and b

4. Thanks Flyingsquirrel: your explanation for Problem 147 helped. I understand the solution now.

Sean, thanks for your help on 154. I understand everything you did but I've run into a problem. Each time I solve simultaneously, I get $a=\frac{-8}{3} , b=\frac{2}{3}$ which works fine for $2+a+b=0$ but yields a very strange, nonzero value indeed for $10+4a+b$. I've tried a few times and get the same values every time.

Here's my working:

$2+a+b=0$

$a=-2-b$

$10+4(-2-b)+b=0$

$
b=\frac{2}{3}$
and $a=-2-\frac{2}{3}=\frac{-8}{3}$

Does this happen for you as well? Or am I erring?

5. Originally Posted by jerry
Thanks Flyingsquirrel: your explanation for Problem 147 helped. I understand the solution now.

Sean, thanks for your help on 154. I understand everything you did but I've run into a problem. Each time I solve simultaneously, I get $a=\frac{-8}{3} , b=\frac{2}{3}$ which works fine for $2+a+b=0$ but yields a very strange, nonzero value indeed for $10+4a+b$. I've tried a few times and get the same values every time.

Here's my working:

$2+a+b=0$

$a=-2-b$

$10+4(-2-b)+b=0$

$
b=\frac{2}{3}$
and $a=-2-\frac{2}{3}=\frac{-8}{3}$

Does this happen for you as well? Or am I erring?
The solution works fine!

$10+4a+b$:

$10 + 4 \left( \frac{-8}{3}\right) + \frac{2}{3} = 10 - \frac{32}{3} + \frac{2}{3} = 10 - \frac{30}{3} = 10 - 10 = 0$, as it should .....!!

6. It appears we're done here. Thanks everybody!