How to go from polynomial of third degree to multiplication of two smaller polynomials
$begingroup$
I have a basic calculus question which I should be able to do easily but I just can't remember how to tackle it. I'm working on my linear algebra exam and trying to find the eigenvalues for a certain matrix A. I got the following as the determinant:
$-lambda^{3}+3lambda^{2}+9lambda-27$ and I want to get this in the form of $(lambda+x_{1})^{2}(lambda+x_{2})$. How to get there?
PS: the solution for this particular one is $x_{1}=-3$ and $x_{2}=3$, it case that might help you.
algebra-precalculus polynomials
edited Dec 18 '18 at 12:57
Jyrki Lahtonen
110k13171385
asked Dec 18 '18 at 11:56
MathbeginnerMathbeginner
1908
$endgroup$
add a comment |
$begingroup$
I have a basic calculus question which I should be able to do easily but I just can't remember how to tackle it. I'm working on my linear algebra exam and trying to find the eigenvalues for a certain matrix A. I got the following as the determinant:
$-lambda^{3}+3lambda^{2}+9lambda-27$ and I want to get this in the form of $(lambda+x_{1})^{2}(lambda+x_{2})$. How to get there?
PS: the solution for this particular one is $x_{1}=-3$ and $x_{2}=3$, it case that might help you.
algebra-precalculus polynomials
edited Dec 18 '18 at 12:57
Jyrki Lahtonen
110k13171385
asked Dec 18 '18 at 11:56
MathbeginnerMathbeginner
1908
$endgroup$
1
$begingroup$
The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
$endgroup$
– user376343
Dec 18 '18 at 13:00
$begingroup$
Good point, thanks
$endgroup$
– Mathbeginner
Dec 18 '18 at 14:25
add a comment |
$begingroup$
I have a basic calculus question which I should be able to do easily but I just can't remember how to tackle it. I'm working on my linear algebra exam and trying to find the eigenvalues for a certain matrix A. I got the following as the determinant:
$-lambda^{3}+3lambda^{2}+9lambda-27$ and I want to get this in the form of $(lambda+x_{1})^{2}(lambda+x_{2})$. How to get there?
PS: the solution for this particular one is $x_{1}=-3$ and $x_{2}=3$, it case that might help you.
algebra-precalculus polynomials
edited Dec 18 '18 at 12:57
Jyrki Lahtonen
110k13171385
asked Dec 18 '18 at 11:56
MathbeginnerMathbeginner
1908
$endgroup$
I have a basic calculus question which I should be able to do easily but I just can't remember how to tackle it. I'm working on my linear algebra exam and trying to find the eigenvalues for a certain matrix A. I got the following as the determinant:
$-lambda^{3}+3lambda^{2}+9lambda-27$ and I want to get this in the form of $(lambda+x_{1})^{2}(lambda+x_{2})$. How to get there?
PS: the solution for this particular one is $x_{1}=-3$ and $x_{2}=3$, it case that might help you.
algebra-precalculus polynomials
algebra-precalculus polynomials
edited Dec 18 '18 at 12:57
Jyrki Lahtonen
110k13171385
asked Dec 18 '18 at 11:56
MathbeginnerMathbeginner
1908
edited Dec 18 '18 at 12:57
Jyrki Lahtonen
110k13171385
asked Dec 18 '18 at 11:56
MathbeginnerMathbeginner
1908
edited Dec 18 '18 at 12:57
Jyrki Lahtonen
110k13171385
edited Dec 18 '18 at 12:57
Jyrki Lahtonen
110k13171385
edited Dec 18 '18 at 12:57
Jyrki Lahtonen
110k13171385
110k13171385
asked Dec 18 '18 at 11:56
MathbeginnerMathbeginner
1908
asked Dec 18 '18 at 11:56
MathbeginnerMathbeginner
1908
asked Dec 18 '18 at 11:56
MathbeginnerMathbeginner
1908
1908
1
$begingroup$
The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
$endgroup$
– user376343
Dec 18 '18 at 13:00
$begingroup$
Good point, thanks
$endgroup$
– Mathbeginner
Dec 18 '18 at 14:25
add a comment |
1
$begingroup$
The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
$endgroup$
– user376343
Dec 18 '18 at 13:00
$begingroup$
Good point, thanks
$endgroup$
– Mathbeginner
Dec 18 '18 at 14:25
1
1
$begingroup$
The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
$endgroup$
– user376343
Dec 18 '18 at 13:00
$begingroup$
The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
$endgroup$
– user376343
Dec 18 '18 at 13:00
$begingroup$
Good point, thanks
$endgroup$
– Mathbeginner
Dec 18 '18 at 14:25
$begingroup$
Good point, thanks
$endgroup$
– Mathbeginner
Dec 18 '18 at 14:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using the rational root theorem, you can find those roots that you mentioned. They are $pm3$, as you know. So, divide your original polynomial by $lambda-3$ and then divide what you got by $lambda+3$.
answered Dec 18 '18 at 12:00
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
The polynomial can be factored by grouping:
$$
-lambda^{3}+3lambda^{2}+9lambda-27
= -lambda^{2}(lambda-3)+9(lambda-3)
= -(lambda^{2}-9)(lambda-3)
= -(lambda+3)(lambda-3)^{2}
$$
answered Dec 18 '18 at 14:06
lhflhf
166k10171400
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045077%2fhow-to-go-from-polynomial-of-third-degree-to-multiplication-of-two-smaller-polyn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the rational root theorem, you can find those roots that you mentioned. They are $pm3$, as you know. So, divide your original polynomial by $lambda-3$ and then divide what you got by $lambda+3$.
answered Dec 18 '18 at 12:00
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
Using the rational root theorem, you can find those roots that you mentioned. They are $pm3$, as you know. So, divide your original polynomial by $lambda-3$ and then divide what you got by $lambda+3$.
answered Dec 18 '18 at 12:00
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
Using the rational root theorem, you can find those roots that you mentioned. They are $pm3$, as you know. So, divide your original polynomial by $lambda-3$ and then divide what you got by $lambda+3$.
answered Dec 18 '18 at 12:00
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
Using the rational root theorem, you can find those roots that you mentioned. They are $pm3$, as you know. So, divide your original polynomial by $lambda-3$ and then divide what you got by $lambda+3$.
answered Dec 18 '18 at 12:00
José Carlos SantosJosé Carlos Santos
168k22132236
answered Dec 18 '18 at 12:00
José Carlos SantosJosé Carlos Santos
168k22132236
answered Dec 18 '18 at 12:00
José Carlos SantosJosé Carlos Santos
168k22132236
answered Dec 18 '18 at 12:00
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
add a comment |
add a comment |
$begingroup$
The polynomial can be factored by grouping:
$$
-lambda^{3}+3lambda^{2}+9lambda-27
= -lambda^{2}(lambda-3)+9(lambda-3)
= -(lambda^{2}-9)(lambda-3)
= -(lambda+3)(lambda-3)^{2}
$$
answered Dec 18 '18 at 14:06
lhflhf
166k10171400
$endgroup$
add a comment |
$begingroup$
The polynomial can be factored by grouping:
$$
-lambda^{3}+3lambda^{2}+9lambda-27
= -lambda^{2}(lambda-3)+9(lambda-3)
= -(lambda^{2}-9)(lambda-3)
= -(lambda+3)(lambda-3)^{2}
$$
answered Dec 18 '18 at 14:06
lhflhf
166k10171400
$endgroup$
add a comment |
$begingroup$
The polynomial can be factored by grouping:
$$
-lambda^{3}+3lambda^{2}+9lambda-27
= -lambda^{2}(lambda-3)+9(lambda-3)
= -(lambda^{2}-9)(lambda-3)
= -(lambda+3)(lambda-3)^{2}
$$
answered Dec 18 '18 at 14:06
lhflhf
166k10171400
$endgroup$
The polynomial can be factored by grouping:
$$
-lambda^{3}+3lambda^{2}+9lambda-27
= -lambda^{2}(lambda-3)+9(lambda-3)
= -(lambda^{2}-9)(lambda-3)
= -(lambda+3)(lambda-3)^{2}
$$
answered Dec 18 '18 at 14:06
lhflhf
166k10171400
edited Dec 19 '18 at 11:03
answered Dec 18 '18 at 14:06
lhflhf
166k10171400
answered Dec 18 '18 at 14:06
lhflhf
166k10171400
answered Dec 18 '18 at 14:06
lhflhf
166k10171400
166k10171400
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045077%2fhow-to-go-from-polynomial-of-third-degree-to-multiplication-of-two-smaller-polyn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
$endgroup$
– user376343
Dec 18 '18 at 13:00
$begingroup$
Good point, thanks
$endgroup$
– Mathbeginner
Dec 18 '18 at 14:25