Polynomial Vocabulary: Difficult Problems with Solutions

Solution:
Expand [tex]f(x+k)[/tex] and combine like terms:
[tex]f(x+k)=(x+k)^3+(x+k)+2=x^3+3kx^2+(3k^2+1)x+k^3+k+2[/tex]
Its constant term is [tex]k^3+k+2[/tex] (the value at [tex]x=0[/tex], i.e. [tex]f(k)[/tex]). It is zero when [tex]k^3+k+2=0[/tex]. Factor the left-hand side:
[tex]k^3+k+2=(k^3+1)+(k+1)=(k+1)(k^2-k+1)+(k+1)=(k+1)(k^2-k+2)[/tex]
The second factor is always positive, because
[tex]k^2-k+2=\left(k-\frac{1}{2}\right)^2+\frac{7}{4}\gt 0[/tex]
Therefore the only value is [tex]k=-1[/tex].

Solution:
If [tex]P(x)=a_nx^n+\ldots+a_1x+a_0[/tex], then the value at [tex]x=1[/tex] is the sum of the coefficients, because every power of [tex]1[/tex] equals [tex]1[/tex]:
[tex]P(1)=a_n+\ldots+a_1+a_0[/tex]
So there is no need to expand: it is enough to substitute [tex]x=1[/tex] in the given expression.
[tex]P(1)=(6-5+4-3+1-2)^{2011}=1^{2011}=1[/tex]
The sum of the coefficients is [tex]1[/tex].

Solution:
Let [tex]P(x)=px^2+qx+r[/tex] with [tex]p\ne 0[/tex]. Then
[tex]P(x+1)-P(x)=p\left((x+1)^2-x^2\right)+q\left((x+1)-x\right)=2px+(p+q)[/tex]
This must be identically equal to [tex]8x+3[/tex], so the coefficients of like powers of [tex]x[/tex] coincide:
[tex]2p=8 , \quad p+q=3\ \Rightarrow\ p=4 , \quad q=-1[/tex]
The constant term is [tex]r=P(0)=5[/tex]. Hence
[tex]P(x)=4x^2-x+5[/tex]
Check:
[tex]4(x+1)^2-(x+1)+5-(4x^2-x+5)=8x+4-1=8x+3[/tex]

Solution:
Put [tex]t=x-1[/tex], i.e. [tex]x=t+1[/tex], and write [tex]f[/tex] in powers of [tex]t[/tex]:
[tex]f(t+1)=(t+1)^3+(t+1)^2-(t+1)+1=[/tex]
[tex]=\left(t^3+3t^2+3t+1\right)+\left(t^2+2t+1\right)-t-1+1=t^3+4t^2+4t+2[/tex]
Returning to [tex]x[/tex]:
[tex]f(x)=(x-1)^3+4(x-1)^2+4(x-1)+2[/tex]
Two polynomials in [tex]t[/tex] are equal only when their coefficients are equal, so [tex]a=1[/tex], [tex]b=4[/tex], [tex]c=4[/tex], [tex]d=2[/tex].
Check: [tex]x=1[/tex] gives [tex]d=f(1)=2[/tex], and [tex]x=2[/tex] gives [tex]a+b+c+d=f(2)=11[/tex].