Finding coefficients in polynomials

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Find $p$ and $q$, if $x^3-2x^2+p+q$ is divided by $x^2+x-2$.
I have found $x =-2$ and $x=1$ however when placed into polynomial I am unable to find a value for $p$ and $q$.

edited 50 mins ago

greedoid

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asked 1 hour ago

Sherma

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I suspect that you mean $x^3-2x^2+px+q$. Otherwise there is no need for $q$.
â€“Â Dietrich Burde
1 hour ago

Hint: Compute $left(x^2+x-2right)(x-3)$
â€“Â robjohnâ™¦
9 mins ago

add a commentÂ |Â

up vote
3
down vote

favorite

Find $p$ and $q$, if $x^3-2x^2+p+q$ is divided by $x^2+x-2$.
I have found $x =-2$ and $x=1$ however when placed into polynomial I am unable to find a value for $p$ and $q$.

edited 50 mins ago

greedoid

29.5k93879

asked 1 hour ago

Sherma

191

New contributor

I suspect that you mean $x^3-2x^2+px+q$. Otherwise there is no need for $q$.
â€“Â Dietrich Burde
1 hour ago

Hint: Compute $left(x^2+x-2right)(x-3)$
â€“Â robjohnâ™¦
9 mins ago

add a commentÂ |Â

up vote
3
down vote

favorite

Find $p$ and $q$, if $x^3-2x^2+p+q$ is divided by $x^2+x-2$.
I have found $x =-2$ and $x=1$ however when placed into polynomial I am unable to find a value for $p$ and $q$.

edited 50 mins ago

greedoid

29.5k93879

asked 1 hour ago

Sherma

191

New contributor

Find $p$ and $q$, if $x^3-2x^2+p+q$ is divided by $x^2+x-2$.
I have found $x =-2$ and $x=1$ however when placed into polynomial I am unable to find a value for $p$ and $q$.

algebra-precalculus polynomials

edited 50 mins ago

greedoid

29.5k93879

asked 1 hour ago

Sherma

191

New contributor

edited 50 mins ago

greedoid

29.5k93879

asked 1 hour ago

Sherma

191

New contributor

edited 50 mins ago

greedoid

29.5k93879

edited 50 mins ago

greedoid

29.5k93879

edited 50 mins ago

greedoid

29.5k93879

asked 1 hour ago

Sherma

191

New contributor

asked 1 hour ago

Sherma

191

asked 1 hour ago

Sherma

191

New contributor

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I suspect that you mean $x^3-2x^2+px+q$. Otherwise there is no need for $q$.
â€“Â Dietrich Burde
1 hour ago

Hint: Compute $left(x^2+x-2right)(x-3)$
â€“Â robjohnâ™¦
9 mins ago

add a commentÂ |Â

I suspect that you mean $x^3-2x^2+px+q$. Otherwise there is no need for $q$.
â€“Â Dietrich Burde
1 hour ago

Hint: Compute $left(x^2+x-2right)(x-3)$
â€“Â robjohnâ™¦
9 mins ago

I suspect that you mean $x^3-2x^2+px+q$. Otherwise there is no need for $q$.
â€“Â Dietrich Burde
1 hour ago

Hint: Compute $left(x^2+x-2right)(x-3)$
â€“Â robjohnâ™¦
9 mins ago

add a commentÂ |Â

4 Answers
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up vote
2
down vote

Hint: Multiply out $(x^2+x-2)(x-a)=x^3-2x^2+px+q$ and compare coefficients.

For the polynomial $x^3-2x^2+px+q$ the comparison gives a solution, namely $a=3$, $p=-5$ and $q=6$. For the above polynomial (with typo) there is no solution.

edited 1 hour ago

answered 1 hour ago

Dietrich Burde

75.3k64185

Thanks much "no solution". Maybe teacher had a typo in posting the question. Thanks much
â€“Â Sherma
56 mins ago

@Sherma You could just correct the typo in your question, then it works.
â€“Â Dietrich Burde
43 mins ago

add a commentÂ |Â

up vote
2
down vote

Hint: If $x-a$ divide $p(x)$ then $p(a)=0$, so

since $x^2+x-2=(x+2)(x-1)$ divide $p(x)= x^3-2x^2+p+q$ we have $p(-2)=0$ and $p(1)=0$

edited 59 mins ago

answered 1 hour ago

greedoid

29.5k93879

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up vote
2
down vote

Or you can use Vieta formulas: You find $x_1=-2$ and $x_2=1$ so, since $$x_1+x_2+x_3 = 2implies x_3= 3$$

so $$p = x_1x_2+x_2x_3+x_3x_1=...;;;rm and;;;q =-x_1x_2x_3 = 6$$

answered 55 mins ago

greedoid

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up vote
1
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One can also reduce $x^3 - 2 x^2 + p x + q$ modulo $x^2 + x - 2$ and then demand that the resulting polynomial is identical to zero. This then means that all the coefficients of that polynomial are zero, and that yields the solution. Modulo $q(x) = x^2 + x - 2$ we have:

$$x^2 bmod q(x)= -x + 2$$

and

$$x^3 bmod q(x)= left(-x^2 + 2 xright) bmod q(x) = 3 x-2 $$

Therefore:

$left(x^3 - 2 x^2 + p x + qright) bmod q(x) = (5+p) x + q-6$

which implies that $p = -5$ and $q = 6$.

answered 35 mins ago

Count Iblis

8,01621332

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
2
down vote

Hint: Multiply out $(x^2+x-2)(x-a)=x^3-2x^2+px+q$ and compare coefficients.

For the polynomial $x^3-2x^2+px+q$ the comparison gives a solution, namely $a=3$, $p=-5$ and $q=6$. For the above polynomial (with typo) there is no solution.

edited 1 hour ago

answered 1 hour ago

Dietrich Burde

75.3k64185

Thanks much "no solution". Maybe teacher had a typo in posting the question. Thanks much
â€“Â Sherma
56 mins ago

@Sherma You could just correct the typo in your question, then it works.
â€“Â Dietrich Burde
43 mins ago

add a commentÂ |Â

up vote
2
down vote

Hint: Multiply out $(x^2+x-2)(x-a)=x^3-2x^2+px+q$ and compare coefficients.

For the polynomial $x^3-2x^2+px+q$ the comparison gives a solution, namely $a=3$, $p=-5$ and $q=6$. For the above polynomial (with typo) there is no solution.

edited 1 hour ago

answered 1 hour ago

Dietrich Burde

75.3k64185

Thanks much "no solution". Maybe teacher had a typo in posting the question. Thanks much
â€“Â Sherma
56 mins ago

@Sherma You could just correct the typo in your question, then it works.
â€“Â Dietrich Burde
43 mins ago

add a commentÂ |Â

up vote
2
down vote

Hint: Multiply out $(x^2+x-2)(x-a)=x^3-2x^2+px+q$ and compare coefficients.

For the polynomial $x^3-2x^2+px+q$ the comparison gives a solution, namely $a=3$, $p=-5$ and $q=6$. For the above polynomial (with typo) there is no solution.

edited 1 hour ago

answered 1 hour ago

Dietrich Burde

75.3k64185

Hint: Multiply out $(x^2+x-2)(x-a)=x^3-2x^2+px+q$ and compare coefficients.

For the polynomial $x^3-2x^2+px+q$ the comparison gives a solution, namely $a=3$, $p=-5$ and $q=6$. For the above polynomial (with typo) there is no solution.

edited 1 hour ago

answered 1 hour ago

Dietrich Burde

75.3k64185

edited 1 hour ago

answered 1 hour ago

Dietrich Burde

75.3k64185

answered 1 hour ago

Dietrich Burde

75.3k64185

answered 1 hour ago

Dietrich Burde

75.3k64185

Thanks much "no solution". Maybe teacher had a typo in posting the question. Thanks much
â€“Â Sherma
56 mins ago

@Sherma You could just correct the typo in your question, then it works.
â€“Â Dietrich Burde
43 mins ago

add a commentÂ |Â

Thanks much "no solution". Maybe teacher had a typo in posting the question. Thanks much
â€“Â Sherma
56 mins ago

@Sherma You could just correct the typo in your question, then it works.
â€“Â Dietrich Burde
43 mins ago

Thanks much "no solution". Maybe teacher had a typo in posting the question. Thanks much
â€“Â Sherma
56 mins ago

@Sherma You could just correct the typo in your question, then it works.
â€“Â Dietrich Burde
43 mins ago

add a commentÂ |Â

up vote
2
down vote

Hint: If $x-a$ divide $p(x)$ then $p(a)=0$, so

since $x^2+x-2=(x+2)(x-1)$ divide $p(x)= x^3-2x^2+p+q$ we have $p(-2)=0$ and $p(1)=0$

edited 59 mins ago

answered 1 hour ago

greedoid

29.5k93879

add a commentÂ |Â

up vote
2
down vote

Hint: If $x-a$ divide $p(x)$ then $p(a)=0$, so

since $x^2+x-2=(x+2)(x-1)$ divide $p(x)= x^3-2x^2+p+q$ we have $p(-2)=0$ and $p(1)=0$

edited 59 mins ago

answered 1 hour ago

greedoid

29.5k93879

add a commentÂ |Â

up vote
2
down vote

Hint: If $x-a$ divide $p(x)$ then $p(a)=0$, so

since $x^2+x-2=(x+2)(x-1)$ divide $p(x)= x^3-2x^2+p+q$ we have $p(-2)=0$ and $p(1)=0$

edited 59 mins ago

answered 1 hour ago

greedoid

29.5k93879

Hint: If $x-a$ divide $p(x)$ then $p(a)=0$, so

since $x^2+x-2=(x+2)(x-1)$ divide $p(x)= x^3-2x^2+p+q$ we have $p(-2)=0$ and $p(1)=0$

edited 59 mins ago

answered 1 hour ago

greedoid

29.5k93879

edited 59 mins ago

answered 1 hour ago

greedoid

29.5k93879

answered 1 hour ago

greedoid

29.5k93879

answered 1 hour ago

greedoid

29.5k93879

add a commentÂ |Â

up vote
2
down vote

Or you can use Vieta formulas: You find $x_1=-2$ and $x_2=1$ so, since $$x_1+x_2+x_3 = 2implies x_3= 3$$

so $$p = x_1x_2+x_2x_3+x_3x_1=...;;;rm and;;;q =-x_1x_2x_3 = 6$$

answered 55 mins ago

greedoid

29.5k93879

add a commentÂ |Â

up vote
2
down vote

Or you can use Vieta formulas: You find $x_1=-2$ and $x_2=1$ so, since $$x_1+x_2+x_3 = 2implies x_3= 3$$

so $$p = x_1x_2+x_2x_3+x_3x_1=...;;;rm and;;;q =-x_1x_2x_3 = 6$$

answered 55 mins ago

greedoid

29.5k93879

add a commentÂ |Â

up vote
2
down vote

Or you can use Vieta formulas: You find $x_1=-2$ and $x_2=1$ so, since $$x_1+x_2+x_3 = 2implies x_3= 3$$

so $$p = x_1x_2+x_2x_3+x_3x_1=...;;;rm and;;;q =-x_1x_2x_3 = 6$$

answered 55 mins ago

greedoid

29.5k93879

Or you can use Vieta formulas: You find $x_1=-2$ and $x_2=1$ so, since $$x_1+x_2+x_3 = 2implies x_3= 3$$

so $$p = x_1x_2+x_2x_3+x_3x_1=...;;;rm and;;;q =-x_1x_2x_3 = 6$$

answered 55 mins ago

greedoid

29.5k93879

answered 55 mins ago

greedoid

29.5k93879

answered 55 mins ago

greedoid

29.5k93879

answered 55 mins ago

greedoid

29.5k93879

add a commentÂ |Â

up vote
1
down vote

$$x^2 bmod q(x)= -x + 2$$

and

$$x^3 bmod q(x)= left(-x^2 + 2 xright) bmod q(x) = 3 x-2 $$

Therefore:

$left(x^3 - 2 x^2 + p x + qright) bmod q(x) = (5+p) x + q-6$

which implies that $p = -5$ and $q = 6$.

answered 35 mins ago

Count Iblis

8,01621332

add a commentÂ |Â

up vote
1
down vote

$$x^2 bmod q(x)= -x + 2$$

and

$$x^3 bmod q(x)= left(-x^2 + 2 xright) bmod q(x) = 3 x-2 $$

Therefore:

$left(x^3 - 2 x^2 + p x + qright) bmod q(x) = (5+p) x + q-6$

which implies that $p = -5$ and $q = 6$.

answered 35 mins ago

Count Iblis

8,01621332

add a commentÂ |Â

up vote
1
down vote

$$x^2 bmod q(x)= -x + 2$$

and

$$x^3 bmod q(x)= left(-x^2 + 2 xright) bmod q(x) = 3 x-2 $$

Therefore:

$left(x^3 - 2 x^2 + p x + qright) bmod q(x) = (5+p) x + q-6$

which implies that $p = -5$ and $q = 6$.

answered 35 mins ago

Count Iblis

8,01621332

$$x^2 bmod q(x)= -x + 2$$

and

$$x^3 bmod q(x)= left(-x^2 + 2 xright) bmod q(x) = 3 x-2 $$

Therefore:

$left(x^3 - 2 x^2 + p x + qright) bmod q(x) = (5+p) x + q-6$

which implies that $p = -5$ and $q = 6$.

answered 35 mins ago

Count Iblis

8,01621332

answered 35 mins ago

Count Iblis

8,01621332

answered 35 mins ago

Count Iblis

8,01621332

answered 35 mins ago

Count Iblis

8,01621332

add a commentÂ |Â

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