Online Course Discussion Forum
Math I-C Lecture 9 HW #9.30
How can you do this quickly? Or how do you do this anyways?
Expand $(3x-1)^6$ to get $Ax^6 + Bx^5 + Cx^4 + Dx^3 + Ex^2 + Fx + G$. What is $A+B+C+D+E+F$?
In the polynomial $P(x) = Ax^6 + Bx^5 + Cx^4 + Dx^3 + Ex^2 + Fx + G$, what value of $x$ could you plug in to get the value of $G$? Does this give you an idea of a value of $x$ you could plug in to get $A + B + C + D + G + F + G$?
What do you mean?
Take for example the polynomial $p(x) = (x-3)^3$. If we expand the polynomial we would get $p(x) = x^3 - 9x^2 + 27x - 27$.
If we plug in $x = 0$ we get $p(0) = (0-3)^3 = -27$, which is exactly the value of the constant term of the polynomial. It makes sense we get exactly this, since every term except this one has an $x$. (Note we did not need to expand the polynomial to find this value.)
If our goal was to get instead the sum of all coefficients, that is, $1 - 9 + 27 - 27 = -8$, what value of $x$ should we plug in?
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