1: Suppose f is a polynomial such that f(0) = 47, f(1) = 32, f(2) = -13, and f(3)=16. What is the sum of the coefficients of f?

1:  Suppose f is a polynomial such that f(0) = 47, f(1) = 32, f(2) = -13, and f(3)=16. What is the sum of the coefficients of f?

Let us suppose that we have this form

P(x)  =   ax^3  + bx^2 +  cx  +  d       where d  =  47

So we have that

a  +    b +  c +   47  =  32    ⇒   a +  b +  c  =  -15

8a +  4b  + 2c + 47  = -13  ⇒  8a + 4b + 2c  =  -60

27a + 9b + 3c + 47  =  16  ⇒  27a + 9b + 3c  = -31

Multiply the first equation by -2 and add it to the second equation

Multiply the first equation by -3 and add it to the third equation

6a  +  2b  = -30   ⇒  3a + b =  - 15   

24a + 6b =    14  ⇒ 12a + 3b  = 7

Multiply  the first equation by -3  and add it to the second

3a   = 52   ⇒    a  = 52/3

3(52/3) + b  =  -15  ⇒  52 + b  = -15   ⇒  b =   -67

(52/3) + (-67) + c  =  -15  ⇒  c  = -15 - 52/3 + 67   ⇒  c  =  104/3

So.... the sum of the coefficients is

52/3 - 67 + 104/3   =    -15 

2:  Consider the polynomials [f(x)=4x^3+3x^2+2x+1]and [g(x)=3-4x+5x^2-6x^3.] Find c such that the polynomial f(x)+cg(x) has degree 2.

[4x^3+3x^2+2x+1] + c [ 3-4x+5x^2-6x^3]

[4x^3+3x^2+2x+1]  + c [ -6x^3 + 5x^2 - 4x + 3 ]

For a resulting polynomial to have degree 2,  we must have that

4  - 6c  =  0

4  =  6c

4/6  = c   =  2/3

So  we have the resulting polynomial

[4x^3+3x^2+2x+1]  + (2/3) [ -6x^3 + 5x^2 - 4x + 3 ]  =

(1/3) (19 x^2 - 2 x + 9)

3:  What is the coefficient of x^3 when 7x^4-3x^3 -3x^2-8x + 1 is multiplied by 8x^4+2x^3 - 7x^2 + 3x + 4 and the like terms are combined?

[ 7x^4 - 3x^3 -3x^2 - 8x + 1 ] *  [ 8x^4 + 2x^3 - 7x^2 + 3x + 4 ]

The coefficient of the x^3 term will come from

(-3 * 4)   +  (-3 * 3) + (-8 * -7)  + (1 * 2)  =

-12      -   9          +    56    +   2   =

37