Simplify third degree polynomial equations.
$begingroup$
Given an equation:
$6x^3 - 13x^2 + 8x + 3 = 0$
Broken down to get one form
$(x + {3over2})$
How can you divide the prior equation to know it will simplify to
$(x + {3over2})(6x^2 + 4x + 2) = 0$
algebra-precalculus
$endgroup$
add a comment |
$begingroup$
Given an equation:
$6x^3 - 13x^2 + 8x + 3 = 0$
Broken down to get one form
$(x + {3over2})$
How can you divide the prior equation to know it will simplify to
$(x + {3over2})(6x^2 + 4x + 2) = 0$
algebra-precalculus
$endgroup$
1
$begingroup$
One way is: "long division"
$endgroup$
– coffeemath
Dec 8 '18 at 11:23
$begingroup$
There is one real solution and is negative, but it is not -3/2.
$endgroup$
– user376343
Dec 8 '18 at 11:42
add a comment |
$begingroup$
Given an equation:
$6x^3 - 13x^2 + 8x + 3 = 0$
Broken down to get one form
$(x + {3over2})$
How can you divide the prior equation to know it will simplify to
$(x + {3over2})(6x^2 + 4x + 2) = 0$
algebra-precalculus
$endgroup$
Given an equation:
$6x^3 - 13x^2 + 8x + 3 = 0$
Broken down to get one form
$(x + {3over2})$
How can you divide the prior equation to know it will simplify to
$(x + {3over2})(6x^2 + 4x + 2) = 0$
algebra-precalculus
algebra-precalculus
edited Dec 8 '18 at 11:25
user587054
46511
46511
asked Dec 8 '18 at 11:20
Samuel M.Samuel M.
1204
1204
1
$begingroup$
One way is: "long division"
$endgroup$
– coffeemath
Dec 8 '18 at 11:23
$begingroup$
There is one real solution and is negative, but it is not -3/2.
$endgroup$
– user376343
Dec 8 '18 at 11:42
add a comment |
1
$begingroup$
One way is: "long division"
$endgroup$
– coffeemath
Dec 8 '18 at 11:23
$begingroup$
There is one real solution and is negative, but it is not -3/2.
$endgroup$
– user376343
Dec 8 '18 at 11:42
1
1
$begingroup$
One way is: "long division"
$endgroup$
– coffeemath
Dec 8 '18 at 11:23
$begingroup$
One way is: "long division"
$endgroup$
– coffeemath
Dec 8 '18 at 11:23
$begingroup$
There is one real solution and is negative, but it is not -3/2.
$endgroup$
– user376343
Dec 8 '18 at 11:42
$begingroup$
There is one real solution and is negative, but it is not -3/2.
$endgroup$
– user376343
Dec 8 '18 at 11:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It looks like you have typo. If you take the polynomial $$p(x)=6 x^3 + 13 x^2 + 8 x + 3,$$ then $-3/2$ is a root.
This guarantees that $x+3/2$ is a factor of $p$. This follows easily from the Division Algorithm: if you write $$p=(x+3/2) q(x)+r(x),$$ with $deg r<deg (x+3/2)=1$, then $r$ is constant; from $p(3/2)=0$ it follows that $r=0$. So now you want to find $q$. You can do it by long division, or simply writing $q(x)=ax^2+bx+c$, and then you have
begin{align}
6 x^3 + 13 x^2 + 8 x + 3&=(x+3/2)(ax^2+bx+c)\
&=ax^3+(3a/2+b)x^2+(c+3b/2)x+3c/2.
end{align}
So $3c/2=3$, and $c=2$. Then $2+3b/2=8$, and $b=4$; finally, $3a/2+4=13$, and $a=6$. Thus
$$
q(x)=6x^2+4x+2.
$$
$endgroup$
add a comment |
$begingroup$
Following @MartinArgerami's discovery that there was probably a typo.
if we have a degree of three polynomial we can use the Rational Zero Theorem as such:
Ration Zero Theorem: For $f(x) = a_nx^n + a_{n-1}x^{n-1}+...+a_2x^2 + a_1x + a_0$, every rational zero of $f(x)$, $frac{p}{q}$ has $p$ as a factor of $a_0$ and $q$ as a factor of $a_n$
We have the polynomial $f(x) = 6x^3 + 13x + 8x + 3$
p) Factors of the constant $a_0 = 3$ : $pm 1$, $pm3$
q) Factors of the coefficient $a_n = 6$ : $pm1$, $pm2$, $pm3$, $pm6$
$frac{p}{q_1}$ : $pm1$, $pm3$
$frac{p}{q_2}$ : $pmfrac{1}{2}$, $pmfrac{3}{2}$
$frac{p}{q_3}$ : $pmfrac{1}{3}$, $pmfrac{3}{3}$
$frac{p}{q_4}$ : $pmfrac{1}{6}$, $pmfrac{3}{6}$
We can see that some of these possible zeros are repeated so we can cross some of them out. Our final possible zero list is the following:
$x = pmfrac{1}{6}$, $pmfrac{1}{3}$, $pmfrac{1}{2}$, $pm1$, $pmfrac{3}{2}$, $pm3$
Using synthetic division we try these possible factors until we find one. Let's try $-3$.
When we try $-3$, we are saying, let us see if $(x+3)$ is a factor of $f(x)$
$begin{array}{c|rrr}&6&13&8&3\-3&&-18&15&-69\hline\&6&-5&23&-66\end{array}$
We have a remainder of $-66$ so $-3$ is not a factor. We are looking for a remainder of $0$. For the sake of time, let us try $frac{-3}{2}$ since you have found already that it is a factor.
$begin{array}{c|rrr}&6&13&8&3\-frac{3}{2}&&-9&-6&-3\hline\&6&4&2&0\end{array}$
We now have a $0$ remainder and $-frac{3}{2}$ is a zero.
Thus, $f(x)$ can be written like this:
$f(x) = divisor(x) times quotient(x) + remainder(x) = d(x)q(x) + r(x) \
= (x+frac{3}{2})(6x^2+4x+2)$
you can further factor the quotient to find the remaining zeros of $f(x)$
$endgroup$
add a comment |
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2 Answers
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$begingroup$
It looks like you have typo. If you take the polynomial $$p(x)=6 x^3 + 13 x^2 + 8 x + 3,$$ then $-3/2$ is a root.
This guarantees that $x+3/2$ is a factor of $p$. This follows easily from the Division Algorithm: if you write $$p=(x+3/2) q(x)+r(x),$$ with $deg r<deg (x+3/2)=1$, then $r$ is constant; from $p(3/2)=0$ it follows that $r=0$. So now you want to find $q$. You can do it by long division, or simply writing $q(x)=ax^2+bx+c$, and then you have
begin{align}
6 x^3 + 13 x^2 + 8 x + 3&=(x+3/2)(ax^2+bx+c)\
&=ax^3+(3a/2+b)x^2+(c+3b/2)x+3c/2.
end{align}
So $3c/2=3$, and $c=2$. Then $2+3b/2=8$, and $b=4$; finally, $3a/2+4=13$, and $a=6$. Thus
$$
q(x)=6x^2+4x+2.
$$
$endgroup$
add a comment |
$begingroup$
It looks like you have typo. If you take the polynomial $$p(x)=6 x^3 + 13 x^2 + 8 x + 3,$$ then $-3/2$ is a root.
This guarantees that $x+3/2$ is a factor of $p$. This follows easily from the Division Algorithm: if you write $$p=(x+3/2) q(x)+r(x),$$ with $deg r<deg (x+3/2)=1$, then $r$ is constant; from $p(3/2)=0$ it follows that $r=0$. So now you want to find $q$. You can do it by long division, or simply writing $q(x)=ax^2+bx+c$, and then you have
begin{align}
6 x^3 + 13 x^2 + 8 x + 3&=(x+3/2)(ax^2+bx+c)\
&=ax^3+(3a/2+b)x^2+(c+3b/2)x+3c/2.
end{align}
So $3c/2=3$, and $c=2$. Then $2+3b/2=8$, and $b=4$; finally, $3a/2+4=13$, and $a=6$. Thus
$$
q(x)=6x^2+4x+2.
$$
$endgroup$
add a comment |
$begingroup$
It looks like you have typo. If you take the polynomial $$p(x)=6 x^3 + 13 x^2 + 8 x + 3,$$ then $-3/2$ is a root.
This guarantees that $x+3/2$ is a factor of $p$. This follows easily from the Division Algorithm: if you write $$p=(x+3/2) q(x)+r(x),$$ with $deg r<deg (x+3/2)=1$, then $r$ is constant; from $p(3/2)=0$ it follows that $r=0$. So now you want to find $q$. You can do it by long division, or simply writing $q(x)=ax^2+bx+c$, and then you have
begin{align}
6 x^3 + 13 x^2 + 8 x + 3&=(x+3/2)(ax^2+bx+c)\
&=ax^3+(3a/2+b)x^2+(c+3b/2)x+3c/2.
end{align}
So $3c/2=3$, and $c=2$. Then $2+3b/2=8$, and $b=4$; finally, $3a/2+4=13$, and $a=6$. Thus
$$
q(x)=6x^2+4x+2.
$$
$endgroup$
It looks like you have typo. If you take the polynomial $$p(x)=6 x^3 + 13 x^2 + 8 x + 3,$$ then $-3/2$ is a root.
This guarantees that $x+3/2$ is a factor of $p$. This follows easily from the Division Algorithm: if you write $$p=(x+3/2) q(x)+r(x),$$ with $deg r<deg (x+3/2)=1$, then $r$ is constant; from $p(3/2)=0$ it follows that $r=0$. So now you want to find $q$. You can do it by long division, or simply writing $q(x)=ax^2+bx+c$, and then you have
begin{align}
6 x^3 + 13 x^2 + 8 x + 3&=(x+3/2)(ax^2+bx+c)\
&=ax^3+(3a/2+b)x^2+(c+3b/2)x+3c/2.
end{align}
So $3c/2=3$, and $c=2$. Then $2+3b/2=8$, and $b=4$; finally, $3a/2+4=13$, and $a=6$. Thus
$$
q(x)=6x^2+4x+2.
$$
edited Dec 10 '18 at 18:12
answered Dec 9 '18 at 1:28
Martin ArgeramiMartin Argerami
125k1177177
125k1177177
add a comment |
add a comment |
$begingroup$
Following @MartinArgerami's discovery that there was probably a typo.
if we have a degree of three polynomial we can use the Rational Zero Theorem as such:
Ration Zero Theorem: For $f(x) = a_nx^n + a_{n-1}x^{n-1}+...+a_2x^2 + a_1x + a_0$, every rational zero of $f(x)$, $frac{p}{q}$ has $p$ as a factor of $a_0$ and $q$ as a factor of $a_n$
We have the polynomial $f(x) = 6x^3 + 13x + 8x + 3$
p) Factors of the constant $a_0 = 3$ : $pm 1$, $pm3$
q) Factors of the coefficient $a_n = 6$ : $pm1$, $pm2$, $pm3$, $pm6$
$frac{p}{q_1}$ : $pm1$, $pm3$
$frac{p}{q_2}$ : $pmfrac{1}{2}$, $pmfrac{3}{2}$
$frac{p}{q_3}$ : $pmfrac{1}{3}$, $pmfrac{3}{3}$
$frac{p}{q_4}$ : $pmfrac{1}{6}$, $pmfrac{3}{6}$
We can see that some of these possible zeros are repeated so we can cross some of them out. Our final possible zero list is the following:
$x = pmfrac{1}{6}$, $pmfrac{1}{3}$, $pmfrac{1}{2}$, $pm1$, $pmfrac{3}{2}$, $pm3$
Using synthetic division we try these possible factors until we find one. Let's try $-3$.
When we try $-3$, we are saying, let us see if $(x+3)$ is a factor of $f(x)$
$begin{array}{c|rrr}&6&13&8&3\-3&&-18&15&-69\hline\&6&-5&23&-66\end{array}$
We have a remainder of $-66$ so $-3$ is not a factor. We are looking for a remainder of $0$. For the sake of time, let us try $frac{-3}{2}$ since you have found already that it is a factor.
$begin{array}{c|rrr}&6&13&8&3\-frac{3}{2}&&-9&-6&-3\hline\&6&4&2&0\end{array}$
We now have a $0$ remainder and $-frac{3}{2}$ is a zero.
Thus, $f(x)$ can be written like this:
$f(x) = divisor(x) times quotient(x) + remainder(x) = d(x)q(x) + r(x) \
= (x+frac{3}{2})(6x^2+4x+2)$
you can further factor the quotient to find the remaining zeros of $f(x)$
$endgroup$
add a comment |
$begingroup$
Following @MartinArgerami's discovery that there was probably a typo.
if we have a degree of three polynomial we can use the Rational Zero Theorem as such:
Ration Zero Theorem: For $f(x) = a_nx^n + a_{n-1}x^{n-1}+...+a_2x^2 + a_1x + a_0$, every rational zero of $f(x)$, $frac{p}{q}$ has $p$ as a factor of $a_0$ and $q$ as a factor of $a_n$
We have the polynomial $f(x) = 6x^3 + 13x + 8x + 3$
p) Factors of the constant $a_0 = 3$ : $pm 1$, $pm3$
q) Factors of the coefficient $a_n = 6$ : $pm1$, $pm2$, $pm3$, $pm6$
$frac{p}{q_1}$ : $pm1$, $pm3$
$frac{p}{q_2}$ : $pmfrac{1}{2}$, $pmfrac{3}{2}$
$frac{p}{q_3}$ : $pmfrac{1}{3}$, $pmfrac{3}{3}$
$frac{p}{q_4}$ : $pmfrac{1}{6}$, $pmfrac{3}{6}$
We can see that some of these possible zeros are repeated so we can cross some of them out. Our final possible zero list is the following:
$x = pmfrac{1}{6}$, $pmfrac{1}{3}$, $pmfrac{1}{2}$, $pm1$, $pmfrac{3}{2}$, $pm3$
Using synthetic division we try these possible factors until we find one. Let's try $-3$.
When we try $-3$, we are saying, let us see if $(x+3)$ is a factor of $f(x)$
$begin{array}{c|rrr}&6&13&8&3\-3&&-18&15&-69\hline\&6&-5&23&-66\end{array}$
We have a remainder of $-66$ so $-3$ is not a factor. We are looking for a remainder of $0$. For the sake of time, let us try $frac{-3}{2}$ since you have found already that it is a factor.
$begin{array}{c|rrr}&6&13&8&3\-frac{3}{2}&&-9&-6&-3\hline\&6&4&2&0\end{array}$
We now have a $0$ remainder and $-frac{3}{2}$ is a zero.
Thus, $f(x)$ can be written like this:
$f(x) = divisor(x) times quotient(x) + remainder(x) = d(x)q(x) + r(x) \
= (x+frac{3}{2})(6x^2+4x+2)$
you can further factor the quotient to find the remaining zeros of $f(x)$
$endgroup$
add a comment |
$begingroup$
Following @MartinArgerami's discovery that there was probably a typo.
if we have a degree of three polynomial we can use the Rational Zero Theorem as such:
Ration Zero Theorem: For $f(x) = a_nx^n + a_{n-1}x^{n-1}+...+a_2x^2 + a_1x + a_0$, every rational zero of $f(x)$, $frac{p}{q}$ has $p$ as a factor of $a_0$ and $q$ as a factor of $a_n$
We have the polynomial $f(x) = 6x^3 + 13x + 8x + 3$
p) Factors of the constant $a_0 = 3$ : $pm 1$, $pm3$
q) Factors of the coefficient $a_n = 6$ : $pm1$, $pm2$, $pm3$, $pm6$
$frac{p}{q_1}$ : $pm1$, $pm3$
$frac{p}{q_2}$ : $pmfrac{1}{2}$, $pmfrac{3}{2}$
$frac{p}{q_3}$ : $pmfrac{1}{3}$, $pmfrac{3}{3}$
$frac{p}{q_4}$ : $pmfrac{1}{6}$, $pmfrac{3}{6}$
We can see that some of these possible zeros are repeated so we can cross some of them out. Our final possible zero list is the following:
$x = pmfrac{1}{6}$, $pmfrac{1}{3}$, $pmfrac{1}{2}$, $pm1$, $pmfrac{3}{2}$, $pm3$
Using synthetic division we try these possible factors until we find one. Let's try $-3$.
When we try $-3$, we are saying, let us see if $(x+3)$ is a factor of $f(x)$
$begin{array}{c|rrr}&6&13&8&3\-3&&-18&15&-69\hline\&6&-5&23&-66\end{array}$
We have a remainder of $-66$ so $-3$ is not a factor. We are looking for a remainder of $0$. For the sake of time, let us try $frac{-3}{2}$ since you have found already that it is a factor.
$begin{array}{c|rrr}&6&13&8&3\-frac{3}{2}&&-9&-6&-3\hline\&6&4&2&0\end{array}$
We now have a $0$ remainder and $-frac{3}{2}$ is a zero.
Thus, $f(x)$ can be written like this:
$f(x) = divisor(x) times quotient(x) + remainder(x) = d(x)q(x) + r(x) \
= (x+frac{3}{2})(6x^2+4x+2)$
you can further factor the quotient to find the remaining zeros of $f(x)$
$endgroup$
Following @MartinArgerami's discovery that there was probably a typo.
if we have a degree of three polynomial we can use the Rational Zero Theorem as such:
Ration Zero Theorem: For $f(x) = a_nx^n + a_{n-1}x^{n-1}+...+a_2x^2 + a_1x + a_0$, every rational zero of $f(x)$, $frac{p}{q}$ has $p$ as a factor of $a_0$ and $q$ as a factor of $a_n$
We have the polynomial $f(x) = 6x^3 + 13x + 8x + 3$
p) Factors of the constant $a_0 = 3$ : $pm 1$, $pm3$
q) Factors of the coefficient $a_n = 6$ : $pm1$, $pm2$, $pm3$, $pm6$
$frac{p}{q_1}$ : $pm1$, $pm3$
$frac{p}{q_2}$ : $pmfrac{1}{2}$, $pmfrac{3}{2}$
$frac{p}{q_3}$ : $pmfrac{1}{3}$, $pmfrac{3}{3}$
$frac{p}{q_4}$ : $pmfrac{1}{6}$, $pmfrac{3}{6}$
We can see that some of these possible zeros are repeated so we can cross some of them out. Our final possible zero list is the following:
$x = pmfrac{1}{6}$, $pmfrac{1}{3}$, $pmfrac{1}{2}$, $pm1$, $pmfrac{3}{2}$, $pm3$
Using synthetic division we try these possible factors until we find one. Let's try $-3$.
When we try $-3$, we are saying, let us see if $(x+3)$ is a factor of $f(x)$
$begin{array}{c|rrr}&6&13&8&3\-3&&-18&15&-69\hline\&6&-5&23&-66\end{array}$
We have a remainder of $-66$ so $-3$ is not a factor. We are looking for a remainder of $0$. For the sake of time, let us try $frac{-3}{2}$ since you have found already that it is a factor.
$begin{array}{c|rrr}&6&13&8&3\-frac{3}{2}&&-9&-6&-3\hline\&6&4&2&0\end{array}$
We now have a $0$ remainder and $-frac{3}{2}$ is a zero.
Thus, $f(x)$ can be written like this:
$f(x) = divisor(x) times quotient(x) + remainder(x) = d(x)q(x) + r(x) \
= (x+frac{3}{2})(6x^2+4x+2)$
you can further factor the quotient to find the remaining zeros of $f(x)$
edited Dec 10 '18 at 16:28
answered Dec 9 '18 at 2:32
BucephalusBucephalus
660417
660417
add a comment |
add a comment |
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1
$begingroup$
One way is: "long division"
$endgroup$
– coffeemath
Dec 8 '18 at 11:23
$begingroup$
There is one real solution and is negative, but it is not -3/2.
$endgroup$
– user376343
Dec 8 '18 at 11:42