Find p and q of this polynomial division [duplicate]
$begingroup$
This question already has an answer here:
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$
9 answers
Dividing $x^4 + px^3 + qx^2 - 16x -12 \$ by $(x+1)(x+3)$, the remainder is $2x+3$, find $p$ and $q$
I dont know how to solve this, please i need help
regards.
algebra-precalculus polynomials
edited Jan 5 at 19:27
José Carlos Santos
153k22123226
asked Jan 5 at 19:20
englishworkvgsenglishworkvgs
22
$endgroup$
marked as duplicate by Bill Dubuque algebra-precalculus
Users with the algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 5 at 21:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$
9 answers
Dividing $x^4 + px^3 + qx^2 - 16x -12 \$ by $(x+1)(x+3)$, the remainder is $2x+3$, find $p$ and $q$
I dont know how to solve this, please i need help
regards.
algebra-precalculus polynomials
edited Jan 5 at 19:27
José Carlos Santos
153k22123226
asked Jan 5 at 19:20
englishworkvgsenglishworkvgs
22
$endgroup$
marked as duplicate by Bill Dubuque algebra-precalculus
Users with the algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 5 at 21:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$
9 answers
Dividing $x^4 + px^3 + qx^2 - 16x -12 \$ by $(x+1)(x+3)$, the remainder is $2x+3$, find $p$ and $q$
I dont know how to solve this, please i need help
regards.
algebra-precalculus polynomials
edited Jan 5 at 19:27
José Carlos Santos
153k22123226
asked Jan 5 at 19:20
englishworkvgsenglishworkvgs
22
$endgroup$
This question already has an answer here:
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$
9 answers
Dividing $x^4 + px^3 + qx^2 - 16x -12 \$ by $(x+1)(x+3)$, the remainder is $2x+3$, find $p$ and $q$
I dont know how to solve this, please i need help
regards.
This question already has an answer here:
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$
9 answers
algebra-precalculus polynomials
algebra-precalculus polynomials
edited Jan 5 at 19:27
José Carlos Santos
153k22123226
asked Jan 5 at 19:20
englishworkvgsenglishworkvgs
22
edited Jan 5 at 19:27
José Carlos Santos
153k22123226
asked Jan 5 at 19:20
englishworkvgsenglishworkvgs
22
edited Jan 5 at 19:27
José Carlos Santos
153k22123226
edited Jan 5 at 19:27
José Carlos Santos
153k22123226
edited Jan 5 at 19:27
José Carlos Santos
153k22123226
153k22123226
asked Jan 5 at 19:20
englishworkvgsenglishworkvgs
22
asked Jan 5 at 19:20
englishworkvgsenglishworkvgs
22
asked Jan 5 at 19:20
englishworkvgsenglishworkvgs
22
22
marked as duplicate by Bill Dubuque algebra-precalculus
Users with the algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 5 at 21:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque algebra-precalculus
Users with the algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 5 at 21:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First substruct $2x+3$ from $p(x)=x^4 + px^3 + qx^2 - 16x -12 \$
begin{align}
q(x) &= p(x) - (2x+3) \
&= x^4 + px^3 + qx^2 - 16x -12 - 2x -3\
&= x^4 + px^3 + qx^2 -18x -15
end{align}
Now, $(x+1)|q(x)$ and $(x+3)|q(x)$
Calculate $q(-1)=0$ and $q(-3)$ to solve $p$ and $q$
answered Jan 5 at 20:19
kelalakakelalaka
329212
$endgroup$
add a comment |
$begingroup$
You know that, for some quadratic polynomial $P(x)$, you have$$x^4+px^3+qx^2-16x-12=(x+1)(x+3)P(x)+2x+3.$$Which information do you extract from this if you put $x=-1$? And what if you put $x=-3$?
answered Jan 5 at 19:25
José Carlos SantosJosé Carlos Santos
153k22123226
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First substruct $2x+3$ from $p(x)=x^4 + px^3 + qx^2 - 16x -12 \$
begin{align}
q(x) &= p(x) - (2x+3) \
&= x^4 + px^3 + qx^2 - 16x -12 - 2x -3\
&= x^4 + px^3 + qx^2 -18x -15
end{align}
Now, $(x+1)|q(x)$ and $(x+3)|q(x)$
Calculate $q(-1)=0$ and $q(-3)$ to solve $p$ and $q$
answered Jan 5 at 20:19
kelalakakelalaka
329212
$endgroup$
add a comment |
$begingroup$
First substruct $2x+3$ from $p(x)=x^4 + px^3 + qx^2 - 16x -12 \$
begin{align}
q(x) &= p(x) - (2x+3) \
&= x^4 + px^3 + qx^2 - 16x -12 - 2x -3\
&= x^4 + px^3 + qx^2 -18x -15
end{align}
Now, $(x+1)|q(x)$ and $(x+3)|q(x)$
Calculate $q(-1)=0$ and $q(-3)$ to solve $p$ and $q$
answered Jan 5 at 20:19
kelalakakelalaka
329212
$endgroup$
add a comment |
$begingroup$
First substruct $2x+3$ from $p(x)=x^4 + px^3 + qx^2 - 16x -12 \$
begin{align}
q(x) &= p(x) - (2x+3) \
&= x^4 + px^3 + qx^2 - 16x -12 - 2x -3\
&= x^4 + px^3 + qx^2 -18x -15
end{align}
Now, $(x+1)|q(x)$ and $(x+3)|q(x)$
Calculate $q(-1)=0$ and $q(-3)$ to solve $p$ and $q$
answered Jan 5 at 20:19
kelalakakelalaka
329212
$endgroup$
First substruct $2x+3$ from $p(x)=x^4 + px^3 + qx^2 - 16x -12 \$
begin{align}
q(x) &= p(x) - (2x+3) \
&= x^4 + px^3 + qx^2 - 16x -12 - 2x -3\
&= x^4 + px^3 + qx^2 -18x -15
end{align}
Now, $(x+1)|q(x)$ and $(x+3)|q(x)$
Calculate $q(-1)=0$ and $q(-3)$ to solve $p$ and $q$
answered Jan 5 at 20:19
kelalakakelalaka
329212
answered Jan 5 at 20:19
kelalakakelalaka
329212
answered Jan 5 at 20:19
kelalakakelalaka
329212
answered Jan 5 at 20:19
kelalakakelalaka
329212
329212
add a comment |
add a comment |
$begingroup$
You know that, for some quadratic polynomial $P(x)$, you have$$x^4+px^3+qx^2-16x-12=(x+1)(x+3)P(x)+2x+3.$$Which information do you extract from this if you put $x=-1$? And what if you put $x=-3$?
answered Jan 5 at 19:25
José Carlos SantosJosé Carlos Santos
153k22123226
$endgroup$
add a comment |
$begingroup$
You know that, for some quadratic polynomial $P(x)$, you have$$x^4+px^3+qx^2-16x-12=(x+1)(x+3)P(x)+2x+3.$$Which information do you extract from this if you put $x=-1$? And what if you put $x=-3$?
answered Jan 5 at 19:25
José Carlos SantosJosé Carlos Santos
153k22123226
$endgroup$
add a comment |
$begingroup$
You know that, for some quadratic polynomial $P(x)$, you have$$x^4+px^3+qx^2-16x-12=(x+1)(x+3)P(x)+2x+3.$$Which information do you extract from this if you put $x=-1$? And what if you put $x=-3$?
answered Jan 5 at 19:25
José Carlos SantosJosé Carlos Santos
153k22123226
$endgroup$
You know that, for some quadratic polynomial $P(x)$, you have$$x^4+px^3+qx^2-16x-12=(x+1)(x+3)P(x)+2x+3.$$Which information do you extract from this if you put $x=-1$? And what if you put $x=-3$?
answered Jan 5 at 19:25
José Carlos SantosJosé Carlos Santos
153k22123226
edited Jan 6 at 8:01
answered Jan 5 at 19:25
José Carlos SantosJosé Carlos Santos
153k22123226
answered Jan 5 at 19:25
José Carlos SantosJosé Carlos Santos
153k22123226
answered Jan 5 at 19:25
José Carlos SantosJosé Carlos Santos
153k22123226
153k22123226
add a comment |
add a comment |
