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Question Number 110374 by Aina Samuel Temidayo last updated on 28/Aug/20
If P(x) is a polynomial whose sum of coefficients is 3 and P(x) can be factorised into two polynomials Q(x),R(x) with integer coefficients, the sum of the coefficients Q(x)^2 +R(x)^2 is
$$\mathrm{If}\:\mathrm{P}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{whose}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{coefficients}\:\mathrm{is}\:\mathrm{3}\:\mathrm{and}\:\mathrm{P}\left(\mathrm{x}\right)\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{factorised}\:\mathrm{into}\:\mathrm{two}\:\mathrm{polynomials} \\ $$$$\mathrm{Q}\left(\mathrm{x}\right),\mathrm{R}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{integer}\:\mathrm{coefficients}, \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{coefficients} \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)^{\mathrm{2}} +\mathrm{R}\left(\mathrm{x}\right)^{\mathrm{2}} \:\mathrm{is} \\ $$
Answered by mr W last updated on 28/Aug/20
Q(x)=Σ_(i=0) ^m q_i x^i R(x)=Σ_(j=0) ^n r_j x^j P(x)=Q(x)R(x)=(Σ_(i=0) ^m q_i x^i )(Σ_(j=0) ^n r_j x^j ) P(1)=(Σ_(i=0) ^m q_i )(Σ_(j=0) ^n r_j )=3 ⇒Σ_(i=0) ^m q_i =±1, Σ_(j=0) ^n r_j =±3 or Σ_(i=0) ^m q_i =±3, Σ_(j=0) ^n r_j =±1 sum of coef. of Q(x)^2 +R(x)^2 is Q(1)^2 +R(1)^2 =(Σ_(i=0) ^m q_i )^2 +(Σ_(j=0) ^n r_j )^2 =10
$${Q}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{{m}} {\sum}}{q}_{{i}} {x}^{{i}} \\ $$$${R}\left({x}\right)=\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}{r}_{{j}} {x}^{{j}} \\ $$$${P}\left({x}\right)={Q}\left({x}\right){R}\left({x}\right)=\left(\underset{{i}=\mathrm{0}} {\overset{{m}} {\sum}}{q}_{{i}} {x}^{{i}} \right)\left(\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}{r}_{{j}} {x}^{{j}} \right) \\ $$$${P}\left(\mathrm{1}\right)=\left(\underset{{i}=\mathrm{0}} {\overset{{m}} {\sum}}{q}_{{i}} \right)\left(\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}{r}_{{j}} \right)=\mathrm{3} \\ $$$$\Rightarrow\underset{{i}=\mathrm{0}} {\overset{{m}} {\sum}}{q}_{{i}} =\pm\mathrm{1},\:\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}{r}_{{j}} =\pm\mathrm{3}\:{or}\:\underset{{i}=\mathrm{0}} {\overset{{m}} {\sum}}{q}_{{i}} =\pm\mathrm{3},\:\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}{r}_{{j}} =\pm\mathrm{1} \\ $$$$ \\ $$$${sum}\:{of}\:{coef}.\:{of}\:\mathrm{Q}\left(\mathrm{x}\right)^{\mathrm{2}} +\mathrm{R}\left(\mathrm{x}\right)^{\mathrm{2}} \:\mathrm{is} \\ $$$$\mathrm{Q}\left(\mathrm{1}\right)^{\mathrm{2}} +\mathrm{R}\left(\mathrm{1}\right)^{\mathrm{2}} =\left(\underset{{i}=\mathrm{0}} {\overset{{m}} {\sum}}{q}_{{i}} \right)^{\mathrm{2}} +\left(\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}{r}_{{j}} \right)^{\mathrm{2}} =\mathrm{10} \\ $$
Commented by Aina Samuel Temidayo last updated on 28/Aug/20
Thanks.
$$\mathrm{Thanks}. \\ $$
Answered by floor(10²Eta[1]) last updated on 28/Aug/20
P(1)=3 P(x)=Q(x).R(x) F(x)=Q(x)^2 +R(x)^2 F(1)=Q(1)^2 +R(1)^2 Q(x)=((P(x))/(R(x)))∴Q(x)^2 =((P(x)^2 )/(R(x)^2 )) ⇒Q(1)^2 =((R(1)^2 )/(P(1)^2 ))∴Q(1)^2 =((R(1)^2 )/9) F(1)=R(1)^2 .((10)/9) since Q(x) and R(x) ∈ Z[x] ⇒the sum of coefficients of these polynomials also have to be an integer. so 3=R(1).Q(1) ⇒R(1)=3 and Q(1)=1 or R(1)=1 and Q(1)=3 or R(1)=−1 and Q(1)=−3 or R(1)=−3 and Q(1)=−1 ∴F(1)=((10)/9)R(1)^2 =10 or ((10)/9) but F(x) ∈ Z[x] because F(x)=Q(x)^2 +R(x)^2 so F(1)∈Z⇒F(1)=10
$$\mathrm{P}\left(\mathrm{1}\right)=\mathrm{3} \\ $$$$\mathrm{P}\left(\mathrm{x}\right)=\mathrm{Q}\left(\mathrm{x}\right).\mathrm{R}\left(\mathrm{x}\right) \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\mathrm{Q}\left(\mathrm{x}\right)^{\mathrm{2}} +\mathrm{R}\left(\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{F}\left(\mathrm{1}\right)=\mathrm{Q}\left(\mathrm{1}\right)^{\mathrm{2}} +\mathrm{R}\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=\frac{\mathrm{P}\left(\mathrm{x}\right)}{\mathrm{R}\left(\mathrm{x}\right)}\therefore\mathrm{Q}\left(\mathrm{x}\right)^{\mathrm{2}} =\frac{\mathrm{P}\left(\mathrm{x}\right)^{\mathrm{2}} }{\mathrm{R}\left(\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{Q}\left(\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{R}\left(\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{P}\left(\mathrm{1}\right)^{\mathrm{2}} }\therefore\mathrm{Q}\left(\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{R}\left(\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{9}} \\ $$$$\mathrm{F}\left(\mathrm{1}\right)=\mathrm{R}\left(\mathrm{1}\right)^{\mathrm{2}} .\frac{\mathrm{10}}{\mathrm{9}} \\ $$$$\mathrm{since}\:\mathrm{Q}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{R}\left(\mathrm{x}\right)\:\in\:\mathbb{Z}\left[\mathrm{x}\right] \\ $$$$\Rightarrow\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{coefficients}\:\mathrm{of}\:\mathrm{these}\:\mathrm{polynomials} \\ $$$$\mathrm{also}\:\mathrm{have}\:\mathrm{to}\:\mathrm{be}\:\mathrm{an}\:\mathrm{integer}. \\ $$$$\mathrm{so}\:\mathrm{3}=\mathrm{R}\left(\mathrm{1}\right).\mathrm{Q}\left(\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{R}\left(\mathrm{1}\right)=\mathrm{3}\:\mathrm{and}\:\mathrm{Q}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{or}\:\mathrm{R}\left(\mathrm{1}\right)=\mathrm{1}\:\mathrm{and}\:\mathrm{Q}\left(\mathrm{1}\right)=\mathrm{3} \\ $$$$\mathrm{or}\:\mathrm{R}\left(\mathrm{1}\right)=−\mathrm{1}\:\mathrm{and}\:\mathrm{Q}\left(\mathrm{1}\right)=−\mathrm{3} \\ $$$$\mathrm{or}\:\mathrm{R}\left(\mathrm{1}\right)=−\mathrm{3}\:\mathrm{and}\:\mathrm{Q}\left(\mathrm{1}\right)=−\mathrm{1} \\ $$$$\therefore\mathrm{F}\left(\mathrm{1}\right)=\frac{\mathrm{10}}{\mathrm{9}}\mathrm{R}\left(\mathrm{1}\right)^{\mathrm{2}} =\mathrm{10}\:\mathrm{or}\:\frac{\mathrm{10}}{\mathrm{9}} \\ $$$$\mathrm{but}\:\mathrm{F}\left(\mathrm{x}\right)\:\in\:\mathbb{Z}\left[\mathrm{x}\right]\:\mathrm{because}\:\mathrm{F}\left(\mathrm{x}\right)=\mathrm{Q}\left(\mathrm{x}\right)^{\mathrm{2}} +\mathrm{R}\left(\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{F}\left(\mathrm{1}\right)\in\mathbb{Z}\Rightarrow\mathrm{F}\left(\mathrm{1}\right)=\mathrm{10} \\ $$$$ \\ $$
Commented by Aina Samuel Temidayo last updated on 28/Aug/20
Thanks.
$$\mathrm{Thanks}. \\ $$

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