Find the value of $Q(x)$ at $x= -1$, knowing some of its properties.
$begingroup$
Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
$$
frac{Q(2x)}{Q(x+1)}= 8 -frac{56}{x+7},
quad forall x ne -7 text{ and } Q(x+1)neq 0,.
$$
Find $Q(-1)$.
algebra-precalculus polynomials rational-functions
edited Dec 10 '18 at 23:24
Batominovski
1
asked Dec 10 '18 at 22:43
James CaldyJames Caldy
191
$endgroup$
add a comment |
$begingroup$
Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
$$
frac{Q(2x)}{Q(x+1)}= 8 -frac{56}{x+7},
quad forall x ne -7 text{ and } Q(x+1)neq 0,.
$$
Find $Q(-1)$.
algebra-precalculus polynomials rational-functions
edited Dec 10 '18 at 23:24
Batominovski
1
asked Dec 10 '18 at 22:43
James CaldyJames Caldy
191
$endgroup$
add a comment |
$begingroup$
Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
$$
frac{Q(2x)}{Q(x+1)}= 8 -frac{56}{x+7},
quad forall x ne -7 text{ and } Q(x+1)neq 0,.
$$
Find $Q(-1)$.
algebra-precalculus polynomials rational-functions
edited Dec 10 '18 at 23:24
Batominovski
1
asked Dec 10 '18 at 22:43
James CaldyJames Caldy
191
$endgroup$
Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
$$
frac{Q(2x)}{Q(x+1)}= 8 -frac{56}{x+7},
quad forall x ne -7 text{ and } Q(x+1)neq 0,.
$$
Find $Q(-1)$.
algebra-precalculus polynomials rational-functions
algebra-precalculus polynomials rational-functions
edited Dec 10 '18 at 23:24
Batominovski
1
asked Dec 10 '18 at 22:43
James CaldyJames Caldy
191
edited Dec 10 '18 at 23:24
Batominovski
1
asked Dec 10 '18 at 22:43
James CaldyJames Caldy
191
edited Dec 10 '18 at 23:24
Batominovski
1
edited Dec 10 '18 at 23:24
Batominovski
1
edited Dec 10 '18 at 23:24
Batominovski
1
1
asked Dec 10 '18 at 22:43
James CaldyJames Caldy
191
asked Dec 10 '18 at 22:43
James CaldyJames Caldy
191
asked Dec 10 '18 at 22:43
James CaldyJames Caldy
191
191
add a comment |
add a comment |
1 Answer
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$begingroup$
Rewrite the equation as
$$(x+7),Q(2x)=8x,Q(x+1),.tag{*}$$
Thus, $2xmid Q(2x)$ or $xmid Q(x)$, and $x+7mid Q(x+1)$ or $x+6mid Q(x)$. That is, $Q(x)=x(x+6),R(x)$ for some polynomial $R(x)$. Show that
$$(x+3),R(2x)=2(x+1),R(x+1),.$$
Ergo, $2x+2mid R(2x)$ or $x+2mid R(x)$, and $x+3mid R(x+1)$ or $x+2mid R(x)$. Thence, $R(x)=(x+2),S(x)$ for some polynomial $S(x)$. Then, prove that
$$S(2x)=S(x+1),,$$
which implies that $S(x)$ is a constant polynomial. Therefore,
$$Q(x)=kx(x+2)(x+6)$$
for some constant $k$.
You can also notice that $limlimits_{xtoinfty},dfrac{Q(2x)}{Q(x+1)}=8=left(dfrac{2}{1}right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0,,$$ and so on.
answered Dec 10 '18 at 23:13
BatominovskiBatominovski
1
$endgroup$
add a comment |
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$begingroup$
Rewrite the equation as
$$(x+7),Q(2x)=8x,Q(x+1),.tag{*}$$
Thus, $2xmid Q(2x)$ or $xmid Q(x)$, and $x+7mid Q(x+1)$ or $x+6mid Q(x)$. That is, $Q(x)=x(x+6),R(x)$ for some polynomial $R(x)$. Show that
$$(x+3),R(2x)=2(x+1),R(x+1),.$$
Ergo, $2x+2mid R(2x)$ or $x+2mid R(x)$, and $x+3mid R(x+1)$ or $x+2mid R(x)$. Thence, $R(x)=(x+2),S(x)$ for some polynomial $S(x)$. Then, prove that
$$S(2x)=S(x+1),,$$
which implies that $S(x)$ is a constant polynomial. Therefore,
$$Q(x)=kx(x+2)(x+6)$$
for some constant $k$.
You can also notice that $limlimits_{xtoinfty},dfrac{Q(2x)}{Q(x+1)}=8=left(dfrac{2}{1}right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0,,$$ and so on.
answered Dec 10 '18 at 23:13
BatominovskiBatominovski
1
$endgroup$
add a comment |
$begingroup$
Rewrite the equation as
$$(x+7),Q(2x)=8x,Q(x+1),.tag{*}$$
Thus, $2xmid Q(2x)$ or $xmid Q(x)$, and $x+7mid Q(x+1)$ or $x+6mid Q(x)$. That is, $Q(x)=x(x+6),R(x)$ for some polynomial $R(x)$. Show that
$$(x+3),R(2x)=2(x+1),R(x+1),.$$
Ergo, $2x+2mid R(2x)$ or $x+2mid R(x)$, and $x+3mid R(x+1)$ or $x+2mid R(x)$. Thence, $R(x)=(x+2),S(x)$ for some polynomial $S(x)$. Then, prove that
$$S(2x)=S(x+1),,$$
which implies that $S(x)$ is a constant polynomial. Therefore,
$$Q(x)=kx(x+2)(x+6)$$
for some constant $k$.
You can also notice that $limlimits_{xtoinfty},dfrac{Q(2x)}{Q(x+1)}=8=left(dfrac{2}{1}right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0,,$$ and so on.
answered Dec 10 '18 at 23:13
BatominovskiBatominovski
1
$endgroup$
add a comment |
$begingroup$
Rewrite the equation as
$$(x+7),Q(2x)=8x,Q(x+1),.tag{*}$$
Thus, $2xmid Q(2x)$ or $xmid Q(x)$, and $x+7mid Q(x+1)$ or $x+6mid Q(x)$. That is, $Q(x)=x(x+6),R(x)$ for some polynomial $R(x)$. Show that
$$(x+3),R(2x)=2(x+1),R(x+1),.$$
Ergo, $2x+2mid R(2x)$ or $x+2mid R(x)$, and $x+3mid R(x+1)$ or $x+2mid R(x)$. Thence, $R(x)=(x+2),S(x)$ for some polynomial $S(x)$. Then, prove that
$$S(2x)=S(x+1),,$$
which implies that $S(x)$ is a constant polynomial. Therefore,
$$Q(x)=kx(x+2)(x+6)$$
for some constant $k$.
You can also notice that $limlimits_{xtoinfty},dfrac{Q(2x)}{Q(x+1)}=8=left(dfrac{2}{1}right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0,,$$ and so on.
answered Dec 10 '18 at 23:13
BatominovskiBatominovski
1
$endgroup$
Rewrite the equation as
$$(x+7),Q(2x)=8x,Q(x+1),.tag{*}$$
Thus, $2xmid Q(2x)$ or $xmid Q(x)$, and $x+7mid Q(x+1)$ or $x+6mid Q(x)$. That is, $Q(x)=x(x+6),R(x)$ for some polynomial $R(x)$. Show that
$$(x+3),R(2x)=2(x+1),R(x+1),.$$
Ergo, $2x+2mid R(2x)$ or $x+2mid R(x)$, and $x+3mid R(x+1)$ or $x+2mid R(x)$. Thence, $R(x)=(x+2),S(x)$ for some polynomial $S(x)$. Then, prove that
$$S(2x)=S(x+1),,$$
which implies that $S(x)$ is a constant polynomial. Therefore,
$$Q(x)=kx(x+2)(x+6)$$
for some constant $k$.
You can also notice that $limlimits_{xtoinfty},dfrac{Q(2x)}{Q(x+1)}=8=left(dfrac{2}{1}right)^3$, so $Q(x)$ must be cubic. If you have no other ideas, then you can try setting $Q(x)=ax^3+bx^2+cx+d$ for some constants $a,b,c,d$. to see how things work out. There are some easily observable results. Plugging in $x:=0$ into (*), you can see that $Q(0)=0$, so $d=0$. With $x:=-7$ in (*), we have $Q(-6)=0$, making $$a(-6)^3+b(-6)^2+c(-6)=0,,$$ and so on.
answered Dec 10 '18 at 23:13
BatominovskiBatominovski
1
edited Dec 10 '18 at 23:20
answered Dec 10 '18 at 23:13
BatominovskiBatominovski
1
answered Dec 10 '18 at 23:13
BatominovskiBatominovski
1
answered Dec 10 '18 at 23:13
BatominovskiBatominovski
1
1
add a comment |
add a comment |
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