Author Topic: Polynomial (Read 2076 times) Tweet Share

deeian · « **on:** April 15, 2016, 01:54:27 pm »

Please help (i can't upload photos because the image size is too big)
1. show that there are two values of k, one of which is integral, for which (x-k) is a factor of p(x), where p(x)=3x^3 + (k+3)x^2 -(4k^2 +k-7)x-4.
Hence for this integral value of k, reduce p(x) into irreducible factors over the real field.

2. consider two polynomial P(x) and F(x).
when p(x) is divided by x^2 +6x+8 the remainder is 2x-11.
when f(x) is divided by x^2 +6x+8 the remainder is x+4.
with each division the quotient is the same
a) show that P(x) and F(x) must have the same degree.

3. p(x) is divided by (x-a)(x-b) show that a remainder R(x) is obtained. show that the remainder is given by
R(x) = (p(a) - p(b) all over a-b)x + aP(b) -bP(a) all over a-b

RuiAce · « **Reply #1 on:** April 15, 2016, 08:59:19 pm »

EDIT BY JAMON: Hey Deeian! I deleted your post in the Extension thread, RuiAce has given some working below, I hope it helps! Let us know if it doesn't make sense. Next time, just post it in one spot, we promise we'll see it and give you a hand just as quick!

$p(x)=3{ x }^{ 3 }+\left( k+3 \right) { x }^{ 2 }-\left( 4{ k }^{ 2 }+k-7 \right) x-4\\\text{If }(x-k)\text{ factors into }p(x)\text{, then according to the factor theorem }p(k)=0\\\begin{align*}\Leftrightarrow 0&=3{k}^{3}+(k+3){k}^{2}-\left( 4{ k }^{ 2 }+k-7 \right) k-4\\\Leftrightarrow 0&=2{k}^{2}+7k-4\\\Leftrightarrow (2k-1)(k+4)&=0\\\Leftrightarrow k&=\frac{1}{2}\text{ or }k=-4\end{align*}$
$\text{Hence the integral value is }k=4\text{, so when this is the case }\\p(x)=3{x}^{3}-{x}^{2}-53x-4\\\text{But the above assumptions have told us that }(x+4)\text{ is a factor.}\\\text{Hence, using polynomial long division (or any other valid approach it can be shown that}\\p(x)=(x+4)\left(3{x}^{2}-13x-1\right)$

$\text{This question is trivial, as by inspection the two polynomials in question have degree 4. One rigorous way to affirm just the fact the degrees are the same:}\\\text{Since the quotients are the same, let us assume the common quotient is }Q(x)\text{ and now}\\\begin{align*}P(x)&=\left({x}^{2}+6x+8\right)Q(x)+(2x-11)\\F(x)&=\left({x}^{2}+6x+8\right)Q(x)+(x+4)\end{align*}$
$\text{Solving simultaneously by subtraction:}\\\begin{align*}P(x)-F(x)&=x-15\\\Leftrightarrow P(x)&=F(x)+x-15\end{align*}\\\text{Now, observe that }P(x)\text{ and }F(x)\text{ must have degree at LEAST 2.}\\\text{This is because there exists a degree 1 quotient when the polynomial is divided by a degree 2 divisor.}\\\text{Hence, since the polynomials are interrelated by only a degree 1 factor, there exists no contradiction to show that they do not have different degrees.}$

Have a look at how you wrote the question.

A remainder R(x) is obtained in essentially obtained in every possible case. You cannot show that a remainder R(x) is obtained because it is trivial.

Happy Physics Land · « **Reply #2 on:** April 16, 2016, 12:22:33 am »

Quote from: RuiAce on April 15, 2016, 08:59:19 pm

EDIT BY JAMON: Hey Deeian! I deleted your post in the Extension thread, RuiAce has given some working below, I hope it helps! Let us know if it doesn't make sense. Next time, just post it in one spot, we promise we'll see it and give you a hand just as quick!

$p(x)=3{ x }^{ 3 }+\left( k+3 \right) { x }^{ 2 }-\left( 4{ k }^{ 2 }+k-7 \right) x-4\\\text{If }(x-k)\text{ factors into }p(x)\text{, then according to the factor theorem }p(k)=0\\\begin{align*}\Leftrightarrow 0&=3{k}^{3}+(k+3){k}^{2}-\left( 4{ k }^{ 2 }+k-7 \right) k-4\\\Leftrightarrow 0&=2{k}^{2}+7k-4\\\Leftrightarrow (2k-1)(k+4)&=0\\\Leftrightarrow k&=\frac{1}{2}\text{ or }k=-4\end{align*}$
$\text{Hence the integral value is }k=4\text{, so when this is the case }\\p(x)=3{x}^{3}-{x}^{2}-53x-4\\\text{But the above assumptions have told us that }(x+4)\text{ is a factor.}\\\text{Hence, using polynomial long division (or any other valid approach it can be shown that}\\p(x)=(x+4)\left(3{x}^{2}-13x-1\right)$
$\text{This question is trivial, as by inspection the two polynomials in question have degree 4. One rigorous way to affirm just the fact the degrees are the same:}\\\text{Since the quotients are the same, let us assume the common quotient is }Q(x)\text{ and now}\\\begin{align*}P(x)&=\left({x}^{2}+6x+8\right)Q(x)+(2x-11)\\F(x)&=\left({x}^{2}+6x+8\right)Q(x)+(x+4)\end{align*}$
$\text{Solving simultaneously by subtraction:}\\\begin{align*}P(x)-F(x)&=x-15\\\Leftrightarrow P(x)&=F(x)+x-15\end{align*}\\\text{Now, observe that }P(x)\text{ and }F(x)\text{ must have degree at LEAST 2.}\\\text{This is because there exists a degree 1 quotient when the polynomial is divided by a degree 2 divisor.}\\\text{Hence, since the polynomials are interrelated by only a degree 1 factor, there exists no contradiction to show that they do not have different degrees.}$
Have a look at how you wrote the question.

A remainder R(x) is obtained in essentially obtained in every possible case. You cannot show that a remainder R(x) is obtained because it is trivial.

The working solution is absolutely grand

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