Finding remainder when unknown $f(x)$ is divided to $g(x)$
$begingroup$
When $f(x)$ is divided by $x - 2$ and $x + 3$, the remainders are 5 and -1, respectively. Find the remainder when $f(x)$ is divided by $x^2 + x - 6$
My method:
Since $x - 2$ and $x + 3$ are linears, dividing by quadratic will leave linear remainders.
Using the remainder theorem:
$$ f(x) = g(x) q(x) + r(x) $$
Where $g(x)$ is a divisor, $q(x)$ is a remainder, and $r(x)$ is a remainder
I let $ax + b$ here be the remainder.
So:
begin{align}
f(x) &= g(x) (x - 2)(x + 3) + ax + b\
f(2) &= g(2) (0)(5) + 2a+ b\
f(-3) &= g(-3) (-5)(0) - 3a + b\
\
&5 = 2a + b\
&-1 = -3a + b\
\
&...\
\
&a = frac{6}{5}, b = frac{13}{5}
\
end{align}
Then the remainder is $ax + b = frac{6}{5}x + frac{13}{5} $
Is it possible to find the $f(x)$ out of remainders?
proof-verification polynomials roots
asked Jan 20 at 8:51
MMJMMMJM
3351111
$endgroup$
add a comment |
$begingroup$
When $f(x)$ is divided by $x - 2$ and $x + 3$, the remainders are 5 and -1, respectively. Find the remainder when $f(x)$ is divided by $x^2 + x - 6$
My method:
Since $x - 2$ and $x + 3$ are linears, dividing by quadratic will leave linear remainders.
Using the remainder theorem:
$$ f(x) = g(x) q(x) + r(x) $$
Where $g(x)$ is a divisor, $q(x)$ is a remainder, and $r(x)$ is a remainder
I let $ax + b$ here be the remainder.
So:
begin{align}
f(x) &= g(x) (x - 2)(x + 3) + ax + b\
f(2) &= g(2) (0)(5) + 2a+ b\
f(-3) &= g(-3) (-5)(0) - 3a + b\
\
&5 = 2a + b\
&-1 = -3a + b\
\
&...\
\
&a = frac{6}{5}, b = frac{13}{5}
\
end{align}
Then the remainder is $ax + b = frac{6}{5}x + frac{13}{5} $
Is it possible to find the $f(x)$ out of remainders?
proof-verification polynomials roots
asked Jan 20 at 8:51
MMJMMMJM
3351111
$endgroup$
add a comment |
$begingroup$
When $f(x)$ is divided by $x - 2$ and $x + 3$, the remainders are 5 and -1, respectively. Find the remainder when $f(x)$ is divided by $x^2 + x - 6$
My method:
Since $x - 2$ and $x + 3$ are linears, dividing by quadratic will leave linear remainders.
Using the remainder theorem:
$$ f(x) = g(x) q(x) + r(x) $$
Where $g(x)$ is a divisor, $q(x)$ is a remainder, and $r(x)$ is a remainder
I let $ax + b$ here be the remainder.
So:
begin{align}
f(x) &= g(x) (x - 2)(x + 3) + ax + b\
f(2) &= g(2) (0)(5) + 2a+ b\
f(-3) &= g(-3) (-5)(0) - 3a + b\
\
&5 = 2a + b\
&-1 = -3a + b\
\
&...\
\
&a = frac{6}{5}, b = frac{13}{5}
\
end{align}
Then the remainder is $ax + b = frac{6}{5}x + frac{13}{5} $
Is it possible to find the $f(x)$ out of remainders?
proof-verification polynomials roots
asked Jan 20 at 8:51
MMJMMMJM
3351111
$endgroup$
When $f(x)$ is divided by $x - 2$ and $x + 3$, the remainders are 5 and -1, respectively. Find the remainder when $f(x)$ is divided by $x^2 + x - 6$
My method:
Since $x - 2$ and $x + 3$ are linears, dividing by quadratic will leave linear remainders.
Using the remainder theorem:
$$ f(x) = g(x) q(x) + r(x) $$
Where $g(x)$ is a divisor, $q(x)$ is a remainder, and $r(x)$ is a remainder
I let $ax + b$ here be the remainder.
So:
begin{align}
f(x) &= g(x) (x - 2)(x + 3) + ax + b\
f(2) &= g(2) (0)(5) + 2a+ b\
f(-3) &= g(-3) (-5)(0) - 3a + b\
\
&5 = 2a + b\
&-1 = -3a + b\
\
&...\
\
&a = frac{6}{5}, b = frac{13}{5}
\
end{align}
Then the remainder is $ax + b = frac{6}{5}x + frac{13}{5} $
Is it possible to find the $f(x)$ out of remainders?
proof-verification polynomials roots
proof-verification polynomials roots
asked Jan 20 at 8:51
MMJMMMJM
3351111
asked Jan 20 at 8:51
MMJMMMJM
3351111
edited Jan 20 at 9:33
MMJM
asked Jan 20 at 8:51
MMJMMMJM
3351111
asked Jan 20 at 8:51
MMJMMMJM
3351111
asked Jan 20 at 8:51
MMJMMMJM
3351111
3351111
add a comment |
add a comment |
1 Answer
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$begingroup$
No, there are infinite $f(x)$ which satisfy the given conditions:
$$f(x)=q(x)(x^2 + x - 6)+frac{6}{5}x + frac{13}{5}$$
with any polynomial $q(x)$.
answered Jan 20 at 8:59
Robert ZRobert Z
98.3k1067139
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add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
No, there are infinite $f(x)$ which satisfy the given conditions:
$$f(x)=q(x)(x^2 + x - 6)+frac{6}{5}x + frac{13}{5}$$
with any polynomial $q(x)$.
answered Jan 20 at 8:59
Robert ZRobert Z
98.3k1067139
$endgroup$
add a comment |
$begingroup$
No, there are infinite $f(x)$ which satisfy the given conditions:
$$f(x)=q(x)(x^2 + x - 6)+frac{6}{5}x + frac{13}{5}$$
with any polynomial $q(x)$.
answered Jan 20 at 8:59
Robert ZRobert Z
98.3k1067139
$endgroup$
add a comment |
$begingroup$
No, there are infinite $f(x)$ which satisfy the given conditions:
$$f(x)=q(x)(x^2 + x - 6)+frac{6}{5}x + frac{13}{5}$$
with any polynomial $q(x)$.
answered Jan 20 at 8:59
Robert ZRobert Z
98.3k1067139
$endgroup$
No, there are infinite $f(x)$ which satisfy the given conditions:
$$f(x)=q(x)(x^2 + x - 6)+frac{6}{5}x + frac{13}{5}$$
with any polynomial $q(x)$.
answered Jan 20 at 8:59
Robert ZRobert Z
98.3k1067139
edited Jan 20 at 9:05
answered Jan 20 at 8:59
Robert ZRobert Z
98.3k1067139
answered Jan 20 at 8:59
Robert ZRobert Z
98.3k1067139
answered Jan 20 at 8:59
Robert ZRobert Z
98.3k1067139
98.3k1067139
add a comment |
add a comment |
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