Mixing
Trick to factor: $n^3 + 6n^2 + 9n + 4$
$begingroup$
Is there any way to tell if
$$n^3 + 6n^2 + 9n + 4$$
has a factored (condensed) form, and, if so, what that factored form is? If the answer is yes to both, may one explain the more general principle?
polynomials
edited Feb 2 at 13:47
greedoid
46.3k1160118
asked Jan 21 at 19:47
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
add a comment |
$begingroup$
Is there any way to tell if
$$n^3 + 6n^2 + 9n + 4$$
has a factored (condensed) form, and, if so, what that factored form is? If the answer is yes to both, may one explain the more general principle?
polynomials
edited Feb 2 at 13:47
greedoid
46.3k1160118
asked Jan 21 at 19:47
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
3
$begingroup$
Hint: search for rational roots.
$endgroup$
– lulu
Jan 21 at 19:48
1
$begingroup$
You can't simplyfie it, you can factor it.
$endgroup$
– greedoid
Jan 21 at 19:48
add a comment |
$begingroup$
Is there any way to tell if
$$n^3 + 6n^2 + 9n + 4$$
has a factored (condensed) form, and, if so, what that factored form is? If the answer is yes to both, may one explain the more general principle?
polynomials
edited Feb 2 at 13:47
greedoid
46.3k1160118
asked Jan 21 at 19:47
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
Is there any way to tell if
$$n^3 + 6n^2 + 9n + 4$$
has a factored (condensed) form, and, if so, what that factored form is? If the answer is yes to both, may one explain the more general principle?
polynomials
polynomials
edited Feb 2 at 13:47
greedoid
46.3k1160118
asked Jan 21 at 19:47
Rafael VergnaudRafael Vergnaud
357217
edited Feb 2 at 13:47
greedoid
46.3k1160118
asked Jan 21 at 19:47
Rafael VergnaudRafael Vergnaud
357217
edited Feb 2 at 13:47
greedoid
46.3k1160118
edited Feb 2 at 13:47
greedoid
46.3k1160118
edited Feb 2 at 13:47
greedoid
46.3k1160118
46.3k1160118
asked Jan 21 at 19:47
Rafael VergnaudRafael Vergnaud
357217
asked Jan 21 at 19:47
Rafael VergnaudRafael Vergnaud
357217
asked Jan 21 at 19:47
Rafael VergnaudRafael Vergnaud
357217
357217
3
$begingroup$
Hint: search for rational roots.
$endgroup$
– lulu
Jan 21 at 19:48
1
$begingroup$
You can't simplyfie it, you can factor it.
$endgroup$
– greedoid
Jan 21 at 19:48
add a comment |
3
$begingroup$
Hint: search for rational roots.
$endgroup$
– lulu
Jan 21 at 19:48
1
$begingroup$
You can't simplyfie it, you can factor it.
$endgroup$
– greedoid
Jan 21 at 19:48
3
3
$begingroup$
Hint: search for rational roots.
$endgroup$
– lulu
Jan 21 at 19:48
$begingroup$
Hint: search for rational roots.
$endgroup$
– lulu
Jan 21 at 19:48
1
1
$begingroup$
You can't simplyfie it, you can factor it.
$endgroup$
– greedoid
Jan 21 at 19:48
$begingroup$
You can't simplyfie it, you can factor it.
$endgroup$
– greedoid
Jan 21 at 19:48
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Notice that, for $f(n)=an^3+bn^2+cn^1+dn^0$, we have
$$f(-1)=a-b+c-d=(a+c)-(b+d)$$
In other words, if the sum of the even co-efficients is equal to the sum of the odd co-efficients, $(x+1)$ can be taken out. Here, this holds
Then we have:
$$(n+1)(pn^2+qn+r)=n^3+6n^2+9n+4$$
Expanding out the LHS, we get:
$$pn^3+(p+q)n^2+(q+r)n+r=n^3+6n^2+9n+4$$
Can you see that $p=1, q=5, r=4$ by matching terms?
So we have:
$$n^3+6n^2+9n+4=(n+1)(n^2+5n+4)$$
Factorise the second bracket and you're done.
answered Jan 21 at 20:08
Rhys HughesRhys Hughes
6,9741530
$endgroup$
$begingroup$
Long division is more efficient than indeterminate coefficients.
$endgroup$
– Yves Daoust
Feb 2 at 13:57
add a comment |
$begingroup$
$$ 1+9 = 10 $$
$$ 6+4 = 10 $$
Either $1$ or $-1$ is a root. I'm betting on $-1$
answered Jan 21 at 20:13
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Further $, 2(1+4)=10 $ so $, {-}1,$ is at least a double root, so clearly $, f(n) = (n!+!1)^2(n!+!4) $
$endgroup$
– Bill Dubuque
Jan 21 at 22:26
add a comment |
$begingroup$
Let's factor it!
$$begin{align*}n^3+6n^2+9n+4 & =n^3+6n^2+8n+n+4\ & =n(n^2+6n+8)+n+4\ & =n(n+4)(n+2)+n+4\ & =(n+4)(n^2+2n+1)\ & =(n+1)^2(n+4)end{align*}$$
edited Feb 27 at 17:20
Dietrich Burde
80.3k647104
answered Jan 21 at 20:10
Frank W.Frank W.
3,8001321
$endgroup$
add a comment |
$begingroup$
$$ n^3+6n^2+9n+4 = (n^3+1)+(6n^2+6n)+(3n+3)$$
$$ = (n+1)(n^2-n+1)+6n(n+1)+3(n+1)=$$
$$ = (n+1)(n^2-n+1+6n+3)$$
$$ = (n+1)(n^2+5n+4) =$$ $$= (n+1)^2(n+4)$$
answered Jan 21 at 19:49
greedoidgreedoid
46.3k1160118
$endgroup$
add a comment |
$begingroup$
The general "principle" is called Polynomial factorization. We have
$$
n^3+6n^2+9n+4=(n+1)^2(n+4).
$$
There are many interesting algorithms. In this particular example, it is enough to do the rational root test. It immediately gives that a rational root must be of the form $pm 1,pm 4$, and then you just try. For a more demanding example, then, try to factorize
$$
n^9 + 3n^7 + 3n^6 + 3n^5 + 6n^4 + 4n^3 + 3n^2 + 3n + 1.
$$
The result is very short and nice.
answered Jan 21 at 19:51
Dietrich BurdeDietrich Burde
80.3k647104
$endgroup$
$begingroup$
Thanks, Dietrich! :)
$endgroup$
– Rafael Vergnaud
Jan 21 at 19:53
1
$begingroup$
Surely you mean polynomial factorization???
$endgroup$
– Wojowu
Jan 21 at 19:53
1
$begingroup$
@Wojowu Yes, was a typo, thank you. Corrected now!
$endgroup$
– Dietrich Burde
Jan 21 at 19:54
add a comment |
$begingroup$
By inspection, $1+9=6+4$ so that $-1$ is a root. Now by long division,
$$frac{n^3+6n^2+9n+4}{n+1}=n^2+5n+4$$ and you can use the quadratic formula.
Or by inspection, $-1$ is again a root and
$$frac{n^2+5n+1}{n+1}=n+4.$$
answered Feb 2 at 13:56
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that, for $f(n)=an^3+bn^2+cn^1+dn^0$, we have
$$f(-1)=a-b+c-d=(a+c)-(b+d)$$
In other words, if the sum of the even co-efficients is equal to the sum of the odd co-efficients, $(x+1)$ can be taken out. Here, this holds
Then we have:
$$(n+1)(pn^2+qn+r)=n^3+6n^2+9n+4$$
Expanding out the LHS, we get:
$$pn^3+(p+q)n^2+(q+r)n+r=n^3+6n^2+9n+4$$
Can you see that $p=1, q=5, r=4$ by matching terms?
So we have:
$$n^3+6n^2+9n+4=(n+1)(n^2+5n+4)$$
Factorise the second bracket and you're done.
answered Jan 21 at 20:08
Rhys HughesRhys Hughes
6,9741530
$endgroup$
$begingroup$
Long division is more efficient than indeterminate coefficients.
$endgroup$
– Yves Daoust
Feb 2 at 13:57
add a comment |
$begingroup$
Notice that, for $f(n)=an^3+bn^2+cn^1+dn^0$, we have
$$f(-1)=a-b+c-d=(a+c)-(b+d)$$
In other words, if the sum of the even co-efficients is equal to the sum of the odd co-efficients, $(x+1)$ can be taken out. Here, this holds
Then we have:
$$(n+1)(pn^2+qn+r)=n^3+6n^2+9n+4$$
Expanding out the LHS, we get:
$$pn^3+(p+q)n^2+(q+r)n+r=n^3+6n^2+9n+4$$
Can you see that $p=1, q=5, r=4$ by matching terms?
So we have:
$$n^3+6n^2+9n+4=(n+1)(n^2+5n+4)$$
Factorise the second bracket and you're done.
answered Jan 21 at 20:08
Rhys HughesRhys Hughes
6,9741530
$endgroup$
$begingroup$
Long division is more efficient than indeterminate coefficients.
$endgroup$
– Yves Daoust
Feb 2 at 13:57
add a comment |
$begingroup$
Notice that, for $f(n)=an^3+bn^2+cn^1+dn^0$, we have
$$f(-1)=a-b+c-d=(a+c)-(b+d)$$
In other words, if the sum of the even co-efficients is equal to the sum of the odd co-efficients, $(x+1)$ can be taken out. Here, this holds
Then we have:
$$(n+1)(pn^2+qn+r)=n^3+6n^2+9n+4$$
Expanding out the LHS, we get:
$$pn^3+(p+q)n^2+(q+r)n+r=n^3+6n^2+9n+4$$
Can you see that $p=1, q=5, r=4$ by matching terms?
So we have:
$$n^3+6n^2+9n+4=(n+1)(n^2+5n+4)$$
Factorise the second bracket and you're done.
answered Jan 21 at 20:08
Rhys HughesRhys Hughes
6,9741530
$endgroup$
Notice that, for $f(n)=an^3+bn^2+cn^1+dn^0$, we have
$$f(-1)=a-b+c-d=(a+c)-(b+d)$$
In other words, if the sum of the even co-efficients is equal to the sum of the odd co-efficients, $(x+1)$ can be taken out. Here, this holds
Then we have:
$$(n+1)(pn^2+qn+r)=n^3+6n^2+9n+4$$
Expanding out the LHS, we get:
$$pn^3+(p+q)n^2+(q+r)n+r=n^3+6n^2+9n+4$$
Can you see that $p=1, q=5, r=4$ by matching terms?
So we have:
$$n^3+6n^2+9n+4=(n+1)(n^2+5n+4)$$
Factorise the second bracket and you're done.
answered Jan 21 at 20:08
Rhys HughesRhys Hughes
6,9741530
answered Jan 21 at 20:08
Rhys HughesRhys Hughes
6,9741530
answered Jan 21 at 20:08
Rhys HughesRhys Hughes
6,9741530
answered Jan 21 at 20:08
Rhys HughesRhys Hughes
6,9741530
6,9741530
$begingroup$
Long division is more efficient than indeterminate coefficients.
$endgroup$
– Yves Daoust
Feb 2 at 13:57
add a comment |
$begingroup$
Long division is more efficient than indeterminate coefficients.
$endgroup$
– Yves Daoust
Feb 2 at 13:57
$begingroup$
Long division is more efficient than indeterminate coefficients.
$endgroup$
– Yves Daoust
Feb 2 at 13:57
$begingroup$
Long division is more efficient than indeterminate coefficients.
$endgroup$
– Yves Daoust
Feb 2 at 13:57
add a comment |
$begingroup$
$$ 1+9 = 10 $$
$$ 6+4 = 10 $$
Either $1$ or $-1$ is a root. I'm betting on $-1$
answered Jan 21 at 20:13
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Further $, 2(1+4)=10 $ so $, {-}1,$ is at least a double root, so clearly $, f(n) = (n!+!1)^2(n!+!4) $
$endgroup$
– Bill Dubuque
Jan 21 at 22:26
add a comment |
$begingroup$
$$ 1+9 = 10 $$
$$ 6+4 = 10 $$
Either $1$ or $-1$ is a root. I'm betting on $-1$
answered Jan 21 at 20:13
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
Further $, 2(1+4)=10 $ so $, {-}1,$ is at least a double root, so clearly $, f(n) = (n!+!1)^2(n!+!4) $
$endgroup$
– Bill Dubuque
Jan 21 at 22:26
add a comment |
$begingroup$
$$ 1+9 = 10 $$
$$ 6+4 = 10 $$
Either $1$ or $-1$ is a root. I'm betting on $-1$
answered Jan 21 at 20:13
Will JagyWill Jagy
104k5102201
$endgroup$
$$ 1+9 = 10 $$
$$ 6+4 = 10 $$
Either $1$ or $-1$ is a root. I'm betting on $-1$
answered Jan 21 at 20:13
Will JagyWill Jagy
104k5102201
answered Jan 21 at 20:13
Will JagyWill Jagy
104k5102201
answered Jan 21 at 20:13
Will JagyWill Jagy
104k5102201
answered Jan 21 at 20:13
Will JagyWill Jagy
104k5102201
104k5102201
$begingroup$
Further $, 2(1+4)=10 $ so $, {-}1,$ is at least a double root, so clearly $, f(n) = (n!+!1)^2(n!+!4) $
$endgroup$
– Bill Dubuque
Jan 21 at 22:26
add a comment |
$begingroup$
Further $, 2(1+4)=10 $ so $, {-}1,$ is at least a double root, so clearly $, f(n) = (n!+!1)^2(n!+!4) $
$endgroup$
– Bill Dubuque
Jan 21 at 22:26
$begingroup$
Further $, 2(1+4)=10 $ so $, {-}1,$ is at least a double root, so clearly $, f(n) = (n!+!1)^2(n!+!4) $
$endgroup$
– Bill Dubuque
Jan 21 at 22:26
$begingroup$
Further $, 2(1+4)=10 $ so $, {-}1,$ is at least a double root, so clearly $, f(n) = (n!+!1)^2(n!+!4) $
$endgroup$
– Bill Dubuque
Jan 21 at 22:26
add a comment |
$begingroup$
Let's factor it!
$$begin{align*}n^3+6n^2+9n+4 & =n^3+6n^2+8n+n+4\ & =n(n^2+6n+8)+n+4\ & =n(n+4)(n+2)+n+4\ & =(n+4)(n^2+2n+1)\ & =(n+1)^2(n+4)end{align*}$$
edited Feb 27 at 17:20
Dietrich Burde
80.3k647104
answered Jan 21 at 20:10
Frank W.Frank W.
3,8001321
$endgroup$
add a comment |
$begingroup$
Let's factor it!
$$begin{align*}n^3+6n^2+9n+4 & =n^3+6n^2+8n+n+4\ & =n(n^2+6n+8)+n+4\ & =n(n+4)(n+2)+n+4\ & =(n+4)(n^2+2n+1)\ & =(n+1)^2(n+4)end{align*}$$
edited Feb 27 at 17:20
Dietrich Burde
80.3k647104
answered Jan 21 at 20:10
Frank W.Frank W.
3,8001321
$endgroup$
add a comment |
$begingroup$
Let's factor it!
$$begin{align*}n^3+6n^2+9n+4 & =n^3+6n^2+8n+n+4\ & =n(n^2+6n+8)+n+4\ & =n(n+4)(n+2)+n+4\ & =(n+4)(n^2+2n+1)\ & =(n+1)^2(n+4)end{align*}$$
edited Feb 27 at 17:20
Dietrich Burde
80.3k647104
answered Jan 21 at 20:10
Frank W.Frank W.
3,8001321
$endgroup$
Let's factor it!
$$begin{align*}n^3+6n^2+9n+4 & =n^3+6n^2+8n+n+4\ & =n(n^2+6n+8)+n+4\ & =n(n+4)(n+2)+n+4\ & =(n+4)(n^2+2n+1)\ & =(n+1)^2(n+4)end{align*}$$
edited Feb 27 at 17:20
Dietrich Burde
80.3k647104
answered Jan 21 at 20:10
Frank W.Frank W.
3,8001321
edited Feb 27 at 17:20
Dietrich Burde
80.3k647104
edited Feb 27 at 17:20
Dietrich Burde
80.3k647104
edited Feb 27 at 17:20
Dietrich Burde
80.3k647104
80.3k647104
answered Jan 21 at 20:10
Frank W.Frank W.
3,8001321
answered Jan 21 at 20:10
Frank W.Frank W.
3,8001321
answered Jan 21 at 20:10
Frank W.Frank W.
3,8001321
3,8001321
add a comment |
add a comment |
$begingroup$
$$ n^3+6n^2+9n+4 = (n^3+1)+(6n^2+6n)+(3n+3)$$
$$ = (n+1)(n^2-n+1)+6n(n+1)+3(n+1)=$$
$$ = (n+1)(n^2-n+1+6n+3)$$
$$ = (n+1)(n^2+5n+4) =$$ $$= (n+1)^2(n+4)$$
answered Jan 21 at 19:49
greedoidgreedoid
46.3k1160118
$endgroup$
add a comment |
$begingroup$
$$ n^3+6n^2+9n+4 = (n^3+1)+(6n^2+6n)+(3n+3)$$
$$ = (n+1)(n^2-n+1)+6n(n+1)+3(n+1)=$$
$$ = (n+1)(n^2-n+1+6n+3)$$
$$ = (n+1)(n^2+5n+4) =$$ $$= (n+1)^2(n+4)$$
answered Jan 21 at 19:49
greedoidgreedoid
46.3k1160118
$endgroup$
add a comment |
$begingroup$
$$ n^3+6n^2+9n+4 = (n^3+1)+(6n^2+6n)+(3n+3)$$
$$ = (n+1)(n^2-n+1)+6n(n+1)+3(n+1)=$$
$$ = (n+1)(n^2-n+1+6n+3)$$
$$ = (n+1)(n^2+5n+4) =$$ $$= (n+1)^2(n+4)$$
answered Jan 21 at 19:49
greedoidgreedoid
46.3k1160118
$endgroup$
$$ n^3+6n^2+9n+4 = (n^3+1)+(6n^2+6n)+(3n+3)$$
$$ = (n+1)(n^2-n+1)+6n(n+1)+3(n+1)=$$
$$ = (n+1)(n^2-n+1+6n+3)$$
$$ = (n+1)(n^2+5n+4) =$$ $$= (n+1)^2(n+4)$$
answered Jan 21 at 19:49
greedoidgreedoid
46.3k1160118
answered Jan 21 at 19:49
greedoidgreedoid
46.3k1160118
answered Jan 21 at 19:49
greedoidgreedoid
46.3k1160118
answered Jan 21 at 19:49
greedoidgreedoid
46.3k1160118
46.3k1160118
add a comment |
add a comment |
$begingroup$
The general "principle" is called Polynomial factorization. We have
$$
n^3+6n^2+9n+4=(n+1)^2(n+4).
$$
There are many interesting algorithms. In this particular example, it is enough to do the rational root test. It immediately gives that a rational root must be of the form $pm 1,pm 4$, and then you just try. For a more demanding example, then, try to factorize
$$
n^9 + 3n^7 + 3n^6 + 3n^5 + 6n^4 + 4n^3 + 3n^2 + 3n + 1.
$$
The result is very short and nice.
answered Jan 21 at 19:51
Dietrich BurdeDietrich Burde
80.3k647104
$endgroup$
$begingroup$
Thanks, Dietrich! :)
$endgroup$
– Rafael Vergnaud
Jan 21 at 19:53
1
$begingroup$
Surely you mean polynomial factorization???
$endgroup$
– Wojowu
Jan 21 at 19:53
1
$begingroup$
@Wojowu Yes, was a typo, thank you. Corrected now!
$endgroup$
– Dietrich Burde
Jan 21 at 19:54
add a comment |
$begingroup$
The general "principle" is called Polynomial factorization. We have
$$
n^3+6n^2+9n+4=(n+1)^2(n+4).
$$
There are many interesting algorithms. In this particular example, it is enough to do the rational root test. It immediately gives that a rational root must be of the form $pm 1,pm 4$, and then you just try. For a more demanding example, then, try to factorize
$$
n^9 + 3n^7 + 3n^6 + 3n^5 + 6n^4 + 4n^3 + 3n^2 + 3n + 1.
$$
The result is very short and nice.
answered Jan 21 at 19:51
Dietrich BurdeDietrich Burde
80.3k647104
$endgroup$
$begingroup$
Thanks, Dietrich! :)
$endgroup$
– Rafael Vergnaud
Jan 21 at 19:53
1
$begingroup$
Surely you mean polynomial factorization???
$endgroup$
– Wojowu
Jan 21 at 19:53
1
$begingroup$
@Wojowu Yes, was a typo, thank you. Corrected now!
$endgroup$
– Dietrich Burde
Jan 21 at 19:54
add a comment |
$begingroup$
The general "principle" is called Polynomial factorization. We have
$$
n^3+6n^2+9n+4=(n+1)^2(n+4).
$$
There are many interesting algorithms. In this particular example, it is enough to do the rational root test. It immediately gives that a rational root must be of the form $pm 1,pm 4$, and then you just try. For a more demanding example, then, try to factorize
$$
n^9 + 3n^7 + 3n^6 + 3n^5 + 6n^4 + 4n^3 + 3n^2 + 3n + 1.
$$
The result is very short and nice.
answered Jan 21 at 19:51
Dietrich BurdeDietrich Burde
80.3k647104
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The general "principle" is called Polynomial factorization. We have
$$
n^3+6n^2+9n+4=(n+1)^2(n+4).
$$
There are many interesting algorithms. In this particular example, it is enough to do the rational root test. It immediately gives that a rational root must be of the form $pm 1,pm 4$, and then you just try. For a more demanding example, then, try to factorize
$$
n^9 + 3n^7 + 3n^6 + 3n^5 + 6n^4 + 4n^3 + 3n^2 + 3n + 1.
$$
The result is very short and nice.
answered Jan 21 at 19:51
Dietrich BurdeDietrich Burde
80.3k647104
edited Jan 21 at 20:00
answered Jan 21 at 19:51
Dietrich BurdeDietrich Burde
80.3k647104
answered Jan 21 at 19:51
Dietrich BurdeDietrich Burde
80.3k647104
answered Jan 21 at 19:51
Dietrich BurdeDietrich Burde
80.3k647104
80.3k647104
$begingroup$
Thanks, Dietrich! :)
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– Rafael Vergnaud
Jan 21 at 19:53
1
$begingroup$
Surely you mean polynomial factorization???
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– Wojowu
Jan 21 at 19:53
1
$begingroup$
@Wojowu Yes, was a typo, thank you. Corrected now!
$endgroup$
– Dietrich Burde
Jan 21 at 19:54
add a comment |
$begingroup$
Thanks, Dietrich! :)
$endgroup$
– Rafael Vergnaud
Jan 21 at 19:53
1
$begingroup$
Surely you mean polynomial factorization???
$endgroup$
– Wojowu
Jan 21 at 19:53
1
$begingroup$
@Wojowu Yes, was a typo, thank you. Corrected now!
$endgroup$
– Dietrich Burde
Jan 21 at 19:54
$begingroup$
Thanks, Dietrich! :)
$endgroup$
– Rafael Vergnaud
Jan 21 at 19:53
$begingroup$
Thanks, Dietrich! :)
$endgroup$
– Rafael Vergnaud
Jan 21 at 19:53
1
1
$begingroup$
Surely you mean polynomial factorization???
$endgroup$
– Wojowu
Jan 21 at 19:53
$begingroup$
Surely you mean polynomial factorization???
$endgroup$
– Wojowu
Jan 21 at 19:53
1
1
$begingroup$
@Wojowu Yes, was a typo, thank you. Corrected now!
$endgroup$
– Dietrich Burde
Jan 21 at 19:54
$begingroup$
@Wojowu Yes, was a typo, thank you. Corrected now!
$endgroup$
– Dietrich Burde
Jan 21 at 19:54
add a comment |
$begingroup$
By inspection, $1+9=6+4$ so that $-1$ is a root. Now by long division,
$$frac{n^3+6n^2+9n+4}{n+1}=n^2+5n+4$$ and you can use the quadratic formula.
Or by inspection, $-1$ is again a root and
$$frac{n^2+5n+1}{n+1}=n+4.$$
answered Feb 2 at 13:56
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
By inspection, $1+9=6+4$ so that $-1$ is a root. Now by long division,
$$frac{n^3+6n^2+9n+4}{n+1}=n^2+5n+4$$ and you can use the quadratic formula.
Or by inspection, $-1$ is again a root and
$$frac{n^2+5n+1}{n+1}=n+4.$$
answered Feb 2 at 13:56
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
By inspection, $1+9=6+4$ so that $-1$ is a root. Now by long division,
$$frac{n^3+6n^2+9n+4}{n+1}=n^2+5n+4$$ and you can use the quadratic formula.
Or by inspection, $-1$ is again a root and
$$frac{n^2+5n+1}{n+1}=n+4.$$
answered Feb 2 at 13:56
Yves DaoustYves Daoust
129k676227
$endgroup$
By inspection, $1+9=6+4$ so that $-1$ is a root. Now by long division,
$$frac{n^3+6n^2+9n+4}{n+1}=n^2+5n+4$$ and you can use the quadratic formula.
Or by inspection, $-1$ is again a root and
$$frac{n^2+5n+1}{n+1}=n+4.$$
answered Feb 2 at 13:56
Yves DaoustYves Daoust
129k676227
answered Feb 2 at 13:56
Yves DaoustYves Daoust
129k676227
answered Feb 2 at 13:56
Yves DaoustYves Daoust
129k676227
answered Feb 2 at 13:56
Yves DaoustYves Daoust
129k676227
129k676227
add a comment |
add a comment |
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Hint: search for rational roots.
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– lulu
Jan 21 at 19:48
1
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You can't simplyfie it, you can factor it.
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– greedoid
Jan 21 at 19:48