Mixing
Solution to quadratic and cubic equation with partial root
$begingroup$
I am having trouble understanding how to resolve quadratic and cubic equations using the method described by my university lecturer (I am really interested to know if this method has a name). My calculator gives very different answers to what my lecturer explained and this enquiring mind wants to know why.
The question was
$$x^3+8x^2+16x+3=0$$
A given root was $x=-3$
While my lecturer gives the roots $x_{1,2,3}=1,5,1$.
I'm not fully confident nor conversant with the methodology he employed yet. The general formula need not be related but the steps were something like the following:
$$x^3+8x^2+16x+3=0$$
Where $0=(x+3)(ax^2+bx+c)$ we substitute the original formula values so, expanding the equation to
$$
x^3+8x^2+16x+3=(x+3)(ax^2+bx+c)
$$
And hence expansion to
$$
=>ax^3+3ax^2+bx^2+3bx+cx+3c
$$
Since the first factor is given as $(x+3)$ we need only find the others using any other factorisation method for quadratic equations.
This is where I got very lost.
$$=ax^3+(3a+b)x^2+(3b+c)x+3c$$
By inspection a=1 and c=1 calculated by
$$a: ax^3=x^3$$
$$a: a=1$$
$$b: (3a+b)x^2=8x^2$$
$$b: 3(1)+b=8$$
$$b: b=8-3=5$$
$$c: 3=3c$$
$$c: 1=c$$
So the final factors he explained were $(x+3)(x-1)(x-5)=0$
Is this the correct final answer or is the calculator doing something not explained to me?
Edit: @Gnumbertester states that the roots found are the quadratic equation values of $(x+3)(x^2+5x+1)$
polynomials quadratics cubic-equations
asked Jan 31 at 2:36
RhodieRhodie
1078
$endgroup$
add a comment |
$begingroup$
I am having trouble understanding how to resolve quadratic and cubic equations using the method described by my university lecturer (I am really interested to know if this method has a name). My calculator gives very different answers to what my lecturer explained and this enquiring mind wants to know why.
The question was
$$x^3+8x^2+16x+3=0$$
A given root was $x=-3$
While my lecturer gives the roots $x_{1,2,3}=1,5,1$.
I'm not fully confident nor conversant with the methodology he employed yet. The general formula need not be related but the steps were something like the following:
$$x^3+8x^2+16x+3=0$$
Where $0=(x+3)(ax^2+bx+c)$ we substitute the original formula values so, expanding the equation to
$$
x^3+8x^2+16x+3=(x+3)(ax^2+bx+c)
$$
And hence expansion to
$$
=>ax^3+3ax^2+bx^2+3bx+cx+3c
$$
Since the first factor is given as $(x+3)$ we need only find the others using any other factorisation method for quadratic equations.
This is where I got very lost.
$$=ax^3+(3a+b)x^2+(3b+c)x+3c$$
By inspection a=1 and c=1 calculated by
$$a: ax^3=x^3$$
$$a: a=1$$
$$b: (3a+b)x^2=8x^2$$
$$b: 3(1)+b=8$$
$$b: b=8-3=5$$
$$c: 3=3c$$
$$c: 1=c$$
So the final factors he explained were $(x+3)(x-1)(x-5)=0$
Is this the correct final answer or is the calculator doing something not explained to me?
Edit: @Gnumbertester states that the roots found are the quadratic equation values of $(x+3)(x^2+5x+1)$
polynomials quadratics cubic-equations
asked Jan 31 at 2:36
RhodieRhodie
1078
$endgroup$
$begingroup$
What's a "partial root"?
$endgroup$
– Gnumbertester
Jan 31 at 2:39
$begingroup$
I'm not a mathematician so you'll have to correct me where the terminology is invalid.
$endgroup$
– Rhodie
Jan 31 at 2:40
$begingroup$
This is wrong. It's obvious that $x^3+8x^2+16x+3$ has no positive roots. (If you substitute a positive number for $x,$ each term is positive.) In fact, $x^3+8x^2+16x+3=(x+3)(x^2+5x+1)$
$endgroup$
– saulspatz
Jan 31 at 2:49
add a comment |
$begingroup$
I am having trouble understanding how to resolve quadratic and cubic equations using the method described by my university lecturer (I am really interested to know if this method has a name). My calculator gives very different answers to what my lecturer explained and this enquiring mind wants to know why.
The question was
$$x^3+8x^2+16x+3=0$$
A given root was $x=-3$
While my lecturer gives the roots $x_{1,2,3}=1,5,1$.
I'm not fully confident nor conversant with the methodology he employed yet. The general formula need not be related but the steps were something like the following:
$$x^3+8x^2+16x+3=0$$
Where $0=(x+3)(ax^2+bx+c)$ we substitute the original formula values so, expanding the equation to
$$
x^3+8x^2+16x+3=(x+3)(ax^2+bx+c)
$$
And hence expansion to
$$
=>ax^3+3ax^2+bx^2+3bx+cx+3c
$$
Since the first factor is given as $(x+3)$ we need only find the others using any other factorisation method for quadratic equations.
This is where I got very lost.
$$=ax^3+(3a+b)x^2+(3b+c)x+3c$$
By inspection a=1 and c=1 calculated by
$$a: ax^3=x^3$$
$$a: a=1$$
$$b: (3a+b)x^2=8x^2$$
$$b: 3(1)+b=8$$
$$b: b=8-3=5$$
$$c: 3=3c$$
$$c: 1=c$$
So the final factors he explained were $(x+3)(x-1)(x-5)=0$
Is this the correct final answer or is the calculator doing something not explained to me?
Edit: @Gnumbertester states that the roots found are the quadratic equation values of $(x+3)(x^2+5x+1)$
polynomials quadratics cubic-equations
asked Jan 31 at 2:36
RhodieRhodie
1078
$endgroup$
I am having trouble understanding how to resolve quadratic and cubic equations using the method described by my university lecturer (I am really interested to know if this method has a name). My calculator gives very different answers to what my lecturer explained and this enquiring mind wants to know why.
The question was
$$x^3+8x^2+16x+3=0$$
A given root was $x=-3$
While my lecturer gives the roots $x_{1,2,3}=1,5,1$.
I'm not fully confident nor conversant with the methodology he employed yet. The general formula need not be related but the steps were something like the following:
$$x^3+8x^2+16x+3=0$$
Where $0=(x+3)(ax^2+bx+c)$ we substitute the original formula values so, expanding the equation to
$$
x^3+8x^2+16x+3=(x+3)(ax^2+bx+c)
$$
And hence expansion to
$$
=>ax^3+3ax^2+bx^2+3bx+cx+3c
$$
Since the first factor is given as $(x+3)$ we need only find the others using any other factorisation method for quadratic equations.
This is where I got very lost.
$$=ax^3+(3a+b)x^2+(3b+c)x+3c$$
By inspection a=1 and c=1 calculated by
$$a: ax^3=x^3$$
$$a: a=1$$
$$b: (3a+b)x^2=8x^2$$
$$b: 3(1)+b=8$$
$$b: b=8-3=5$$
$$c: 3=3c$$
$$c: 1=c$$
So the final factors he explained were $(x+3)(x-1)(x-5)=0$
Is this the correct final answer or is the calculator doing something not explained to me?
Edit: @Gnumbertester states that the roots found are the quadratic equation values of $(x+3)(x^2+5x+1)$
polynomials quadratics cubic-equations
polynomials quadratics cubic-equations
asked Jan 31 at 2:36
RhodieRhodie
1078
asked Jan 31 at 2:36
RhodieRhodie
1078
edited Jan 31 at 2:51
Rhodie
asked Jan 31 at 2:36
RhodieRhodie
1078
asked Jan 31 at 2:36
RhodieRhodie
1078
asked Jan 31 at 2:36
RhodieRhodie
1078
1078
$begingroup$
What's a "partial root"?
$endgroup$
– Gnumbertester
Jan 31 at 2:39
$begingroup$
I'm not a mathematician so you'll have to correct me where the terminology is invalid.
$endgroup$
– Rhodie
Jan 31 at 2:40
$begingroup$
This is wrong. It's obvious that $x^3+8x^2+16x+3$ has no positive roots. (If you substitute a positive number for $x,$ each term is positive.) In fact, $x^3+8x^2+16x+3=(x+3)(x^2+5x+1)$
$endgroup$
– saulspatz
Jan 31 at 2:49
add a comment |
$begingroup$
What's a "partial root"?
$endgroup$
– Gnumbertester
Jan 31 at 2:39
$begingroup$
I'm not a mathematician so you'll have to correct me where the terminology is invalid.
$endgroup$
– Rhodie
Jan 31 at 2:40
$begingroup$
This is wrong. It's obvious that $x^3+8x^2+16x+3$ has no positive roots. (If you substitute a positive number for $x,$ each term is positive.) In fact, $x^3+8x^2+16x+3=(x+3)(x^2+5x+1)$
$endgroup$
– saulspatz
Jan 31 at 2:49
$begingroup$
What's a "partial root"?
$endgroup$
– Gnumbertester
Jan 31 at 2:39
$begingroup$
What's a "partial root"?
$endgroup$
– Gnumbertester
Jan 31 at 2:39
$begingroup$
I'm not a mathematician so you'll have to correct me where the terminology is invalid.
$endgroup$
– Rhodie
Jan 31 at 2:40
$begingroup$
I'm not a mathematician so you'll have to correct me where the terminology is invalid.
$endgroup$
– Rhodie
Jan 31 at 2:40
$begingroup$
This is wrong. It's obvious that $x^3+8x^2+16x+3$ has no positive roots. (If you substitute a positive number for $x,$ each term is positive.) In fact, $x^3+8x^2+16x+3=(x+3)(x^2+5x+1)$
$endgroup$
– saulspatz
Jan 31 at 2:49
$begingroup$
This is wrong. It's obvious that $x^3+8x^2+16x+3$ has no positive roots. (If you substitute a positive number for $x,$ each term is positive.) In fact, $x^3+8x^2+16x+3=(x+3)(x^2+5x+1)$
$endgroup$
– saulspatz
Jan 31 at 2:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your answer is incorrect and there is a much simpler way of finding the roots.
Since you are given a root of this cubic, you can use synthetic division to reduce it to a quadratic.
You may already know how to do synthetic division, however if you don't, you can see how to do it here.
Once you divide $x^3+8x^2+16x+3$ by the linear factor you are given, you are left with $x^2+5x+1$.
To find the other two roots, use the quadratic formula.
Addendum: If you are familiar with traditional, polynomial long division, that works too.
answered Jan 31 at 2:46
GnumbertesterGnumbertester
6771114
$endgroup$
$begingroup$
What is a linear factor?
$endgroup$
– Rhodie
Jan 31 at 2:47
$begingroup$
Since you are given the root $x=-3$, you know that $x+3$ is a factor of the cubic. I just included linear because it is a linear term (degree of 1).
$endgroup$
– Gnumbertester
Jan 31 at 2:48
$begingroup$
Above my pay grade buddy...
$endgroup$
– Rhodie
Jan 31 at 2:52
$begingroup$
Linear isn't technically necessary for the understanding of this problem. The most important part is knowing that if you are given a factor of a cubic, you can find an expression (quadratic) by dividing the cubic by that factor. From there, all you need to do is factor the quadratic and you have all 3 roots of the original cubic.
$endgroup$
– Gnumbertester
Jan 31 at 2:55
$begingroup$
You lost me at linear...
$endgroup$
– Rhodie
Jan 31 at 3:15
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%2f3094431%2fsolution-to-quadratic-and-cubic-equation-with-partial-root%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your answer is incorrect and there is a much simpler way of finding the roots.
Since you are given a root of this cubic, you can use synthetic division to reduce it to a quadratic.
You may already know how to do synthetic division, however if you don't, you can see how to do it here.
Once you divide $x^3+8x^2+16x+3$ by the linear factor you are given, you are left with $x^2+5x+1$.
To find the other two roots, use the quadratic formula.
Addendum: If you are familiar with traditional, polynomial long division, that works too.
answered Jan 31 at 2:46
GnumbertesterGnumbertester
6771114
$endgroup$
$begingroup$
What is a linear factor?
$endgroup$
– Rhodie
Jan 31 at 2:47
$begingroup$
Since you are given the root $x=-3$, you know that $x+3$ is a factor of the cubic. I just included linear because it is a linear term (degree of 1).
$endgroup$
– Gnumbertester
Jan 31 at 2:48
$begingroup$
Above my pay grade buddy...
$endgroup$
– Rhodie
Jan 31 at 2:52
$begingroup$
Linear isn't technically necessary for the understanding of this problem. The most important part is knowing that if you are given a factor of a cubic, you can find an expression (quadratic) by dividing the cubic by that factor. From there, all you need to do is factor the quadratic and you have all 3 roots of the original cubic.
$endgroup$
– Gnumbertester
Jan 31 at 2:55
$begingroup$
You lost me at linear...
$endgroup$
– Rhodie
Jan 31 at 3:15
add a comment |
$begingroup$
Your answer is incorrect and there is a much simpler way of finding the roots.
Since you are given a root of this cubic, you can use synthetic division to reduce it to a quadratic.
You may already know how to do synthetic division, however if you don't, you can see how to do it here.
Once you divide $x^3+8x^2+16x+3$ by the linear factor you are given, you are left with $x^2+5x+1$.
To find the other two roots, use the quadratic formula.
Addendum: If you are familiar with traditional, polynomial long division, that works too.
answered Jan 31 at 2:46
GnumbertesterGnumbertester
6771114
$endgroup$
$begingroup$
What is a linear factor?
$endgroup$
– Rhodie
Jan 31 at 2:47
$begingroup$
Since you are given the root $x=-3$, you know that $x+3$ is a factor of the cubic. I just included linear because it is a linear term (degree of 1).
$endgroup$
– Gnumbertester
Jan 31 at 2:48
$begingroup$
Above my pay grade buddy...
$endgroup$
– Rhodie
Jan 31 at 2:52
$begingroup$
Linear isn't technically necessary for the understanding of this problem. The most important part is knowing that if you are given a factor of a cubic, you can find an expression (quadratic) by dividing the cubic by that factor. From there, all you need to do is factor the quadratic and you have all 3 roots of the original cubic.
$endgroup$
– Gnumbertester
Jan 31 at 2:55
$begingroup$
You lost me at linear...
$endgroup$
– Rhodie
Jan 31 at 3:15
add a comment |
$begingroup$
Your answer is incorrect and there is a much simpler way of finding the roots.
Since you are given a root of this cubic, you can use synthetic division to reduce it to a quadratic.
You may already know how to do synthetic division, however if you don't, you can see how to do it here.
Once you divide $x^3+8x^2+16x+3$ by the linear factor you are given, you are left with $x^2+5x+1$.
To find the other two roots, use the quadratic formula.
Addendum: If you are familiar with traditional, polynomial long division, that works too.
answered Jan 31 at 2:46
GnumbertesterGnumbertester
6771114
$endgroup$
Your answer is incorrect and there is a much simpler way of finding the roots.
Since you are given a root of this cubic, you can use synthetic division to reduce it to a quadratic.
You may already know how to do synthetic division, however if you don't, you can see how to do it here.
Once you divide $x^3+8x^2+16x+3$ by the linear factor you are given, you are left with $x^2+5x+1$.
To find the other two roots, use the quadratic formula.
Addendum: If you are familiar with traditional, polynomial long division, that works too.
answered Jan 31 at 2:46
GnumbertesterGnumbertester
6771114
edited Jan 31 at 2:47
answered Jan 31 at 2:46
GnumbertesterGnumbertester
6771114
answered Jan 31 at 2:46
GnumbertesterGnumbertester
6771114
answered Jan 31 at 2:46
GnumbertesterGnumbertester
6771114
6771114
$begingroup$
What is a linear factor?
$endgroup$
– Rhodie
Jan 31 at 2:47
$begingroup$
Since you are given the root $x=-3$, you know that $x+3$ is a factor of the cubic. I just included linear because it is a linear term (degree of 1).
$endgroup$
– Gnumbertester
Jan 31 at 2:48
$begingroup$
Above my pay grade buddy...
$endgroup$
– Rhodie
Jan 31 at 2:52
$begingroup$
Linear isn't technically necessary for the understanding of this problem. The most important part is knowing that if you are given a factor of a cubic, you can find an expression (quadratic) by dividing the cubic by that factor. From there, all you need to do is factor the quadratic and you have all 3 roots of the original cubic.
$endgroup$
– Gnumbertester
Jan 31 at 2:55
$begingroup$
You lost me at linear...
$endgroup$
– Rhodie
Jan 31 at 3:15
add a comment |
$begingroup$
What is a linear factor?
$endgroup$
– Rhodie
Jan 31 at 2:47
$begingroup$
Since you are given the root $x=-3$, you know that $x+3$ is a factor of the cubic. I just included linear because it is a linear term (degree of 1).
$endgroup$
– Gnumbertester
Jan 31 at 2:48
$begingroup$
Above my pay grade buddy...
$endgroup$
– Rhodie
Jan 31 at 2:52
$begingroup$
Linear isn't technically necessary for the understanding of this problem. The most important part is knowing that if you are given a factor of a cubic, you can find an expression (quadratic) by dividing the cubic by that factor. From there, all you need to do is factor the quadratic and you have all 3 roots of the original cubic.
$endgroup$
– Gnumbertester
Jan 31 at 2:55
$begingroup$
You lost me at linear...
$endgroup$
– Rhodie
Jan 31 at 3:15
$begingroup$
What is a linear factor?
$endgroup$
– Rhodie
Jan 31 at 2:47
$begingroup$
What is a linear factor?
$endgroup$
– Rhodie
Jan 31 at 2:47
$begingroup$
Since you are given the root $x=-3$, you know that $x+3$ is a factor of the cubic. I just included linear because it is a linear term (degree of 1).
$endgroup$
– Gnumbertester
Jan 31 at 2:48
$begingroup$
Since you are given the root $x=-3$, you know that $x+3$ is a factor of the cubic. I just included linear because it is a linear term (degree of 1).
$endgroup$
– Gnumbertester
Jan 31 at 2:48
$begingroup$
Above my pay grade buddy...
$endgroup$
– Rhodie
Jan 31 at 2:52
$begingroup$
Above my pay grade buddy...
$endgroup$
– Rhodie
Jan 31 at 2:52
$begingroup$
Linear isn't technically necessary for the understanding of this problem. The most important part is knowing that if you are given a factor of a cubic, you can find an expression (quadratic) by dividing the cubic by that factor. From there, all you need to do is factor the quadratic and you have all 3 roots of the original cubic.
$endgroup$
– Gnumbertester
Jan 31 at 2:55
$begingroup$
Linear isn't technically necessary for the understanding of this problem. The most important part is knowing that if you are given a factor of a cubic, you can find an expression (quadratic) by dividing the cubic by that factor. From there, all you need to do is factor the quadratic and you have all 3 roots of the original cubic.
$endgroup$
– Gnumbertester
Jan 31 at 2:55
$begingroup$
You lost me at linear...
$endgroup$
– Rhodie
Jan 31 at 3:15
$begingroup$
You lost me at linear...
$endgroup$
– Rhodie
Jan 31 at 3:15
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%2f3094431%2fsolution-to-quadratic-and-cubic-equation-with-partial-root%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
$begingroup$
What's a "partial root"?
$endgroup$
– Gnumbertester
Jan 31 at 2:39
$begingroup$
I'm not a mathematician so you'll have to correct me where the terminology is invalid.
$endgroup$
– Rhodie
Jan 31 at 2:40
$begingroup$
This is wrong. It's obvious that $x^3+8x^2+16x+3$ has no positive roots. (If you substitute a positive number for $x,$ each term is positive.) In fact, $x^3+8x^2+16x+3=(x+3)(x^2+5x+1)$
$endgroup$
– saulspatz
Jan 31 at 2:49