Question

Consider a polynomial of degree 4 with leading coefficient 2 such that \[P(1)=1\], \[P(2)=16\], \[P(3)=81\], \[P(4)=256\] then \[P(5)\] is:
A. 673
B. 637
C. 727
D. None of these

Answer 1

Hint: We have to find \[P(5)\] i.e. value of polynomial at \[x=5\], we can assume polynomial to be in the general form as follows \[P(x)=A{{x}^{4}}+B{{x}^{3}}+C{{x}^{2}}+Dx+E\], it is given to us that the leading coefficient is 2 i.e. coefficient of highest degree term in x is 2 (\[A=2\] ), \[P(x)=2{{x}^{4}}+B{{x}^{3}}+C{{x}^{2}}+Dx+E\], Now we have four unknown and four equations which we can use to find the completed equation by equating them in general equation body, thereafter we can find \[P(5)\] by putting \[x=5\].

Complete step-by-step solution:
We assume a general polynomial of degree 4 so that later in the answer we can find the value of unknown constants,
\[P(x)=A{{x}^{4}}+B{{x}^{3}}+C{{x}^{2}}+Dx+E\].
Now, it is given to us in the question that the leading coefficient is 2 i.e. \[A=2\],
\[P(x)=2{{x}^{4}}+B{{x}^{3}}+C{{x}^{2}}+Dx+E\].
The above equation needs to be solved for unknowns,
Now in order to solve any equation, we must have the same number of equations as there are unknowns in the equation, here we have 4 unknowns and 4 equations so we can find unknowns in the above polynomial.
Let us put \[x=1\], since we know that the value of \[P(1)=1\],
\[P(1)={{2.1}^{4}}+B{{.1}^{3}}+C{{.1}^{2}}+D.1+E=1\].
\[\Rightarrow 2+B+C+D+E=1\].
\[\Rightarrow B+C+D+E=-1\ldots eq(1)\].
Similarly, we do so in case of \[x=2\], where \[P(2)=16\],
\[\Rightarrow P(2)=16={{2.2}^{4}}+B{{.2}^{3}}+C{{.2}^{2}}+D.2+E\].
\[\Rightarrow 32+8B+4C+2D+E=16\].
\[\Rightarrow 8B+4C+2D+E=-16\ldots eq(2)\].
Also in the case of \[x=3\] as we know \[P(3)=81\],
\[\Rightarrow P(3)={{2.3}^{4}}+B{{.3}^{3}}+C{{.3}^{2}}+D.3+E=81\].
\[\Rightarrow 162+27B+9C+3D+E=81\].
\[\Rightarrow 27B+9C+3D+E=-81\ldots eq(3)\].
Another given case is \[x=4\] where \[P(4)=256\],
\[P(4)={{2.4}^{4}}+B{{.4}^{3}}+C{{.4}^{2}}+D.4+E=256\].
\[\Rightarrow 512+64B+16C+4D+E=256\].
\[\Rightarrow 64B+16C+4D+E=-256\ldots eq(4)\].
Now we have four equations which can be solved as follows,
Subtracting eq(1) from eq(2) we have
\[7B+3C+D=-15\ldots eq(5)\].
Subtracting eq(2) from eq(3) we have
\[19B+5C+D=-65\ldots eq(6)\].
Subtracting eq(3) from eq(4) we have
\[37B+7C+D=-175\ldots eq(7)\].
Now solving these three equations, since we have three equations and three unknowns, we can solve these equations,
Subtracting eq(5) from eq(6) we have
\[12B+2C=-50\ldots eq(8)\].
Subtracting eq(6) from eq(7) we have
\[18B+2C=-110\ldots eq(9)\].
Subtracting eq(8) from eq(9) we have
\[6B=-60\].
\[\Rightarrow B=-10\].
Using the value of B in eq(8) we get the value of C
We get, \[-120+2C=-50\].
\[\Rightarrow 2C=70\],
\[\Rightarrow C=35\].
Using values of B and C in eq(5)
 we get, \[-70+105+D=-15\].
\[\Rightarrow D=-50\].
Using values of A, B and C in eq(1)
We get
\[-10+35-50+E=-1\].
\[\Rightarrow E=24\].
‘Hence the completed polynomial equation is –
\[P(x)=2{{x}^{4}}-10{{x}^{3}}+35{{x}^{2}}-50x+24\].
Thereafter we can find the answer as follows –
\[\Rightarrow P(5)={{2.5}^{4}}-{{10.5}^{3}}+{{35.5}^{2}}-50.5+24\].
\[\Rightarrow P(5)=1250-1250+875-250+24\].
\[\Rightarrow P(5)=649\]
Therefore the answer is option D

Note: The common mistakes committed by students include wrong substitutions as well as inability to form and analyze the general polynomial equation. The many variables in the core equation must not be confused for one another. The student must understand that in order to find n number of unknowns; we must have the same or more no given relations in order to find the unknown numbers.