2

Question: Determine if the leading coefficient of the function "a", is positive or negative.

a) $$f(x)=(x-3)^2(x+1)(x+2)^3$$

In my notes I stated the sign of the leading coefficient without work but in order to get the answer now I had to expand the polynomial function out. Any help would be appreciated.

-Jack

nonuser
  • 91,557
Jack Curran
  • 23

4 Answers4

2

You asked for the leading coefficient of $$f(x)=(x-3)^2(x+1)(x+2)^3$$

As you see this is a polynomial of degree $6$ therefore you want to know the coefficient of $x^6$

How many $x^6$ are there in the product?

Well there is only $1$ because you need to multiply the leading coefficients of each factor to get the leading coefficient of the product.

In our case all the leading coefficients of factors are $1$ so the leading coefficient of the product is also $1$

Mohammad Riazi-Kermani
  • 69,891
1

In general once we know the roots of a polynomial function in the form

$$f(x)=k(x-x_1)(x-x_2)...(x-x_n)$$

we can easily find all the coefficient for

$$f(x)=a_nx^2+a_{n-1}x^{n-1}+...+a_1x+a_0$$

indeed

$$a_n=k$$

$$a_{n-1}=-(x_1+x_2+...+x_n)$$

$$a_{n-2}=x_1x_2+x_1x_3...+x_{n-1}x_n$$ $$...$$ $$a_0=(-1)^{n}\prod x_i$$

Take also a look here Relation betwen coefficients and roots of a polynomial

user
  • 162,563
0

You could just replace the numbers 3, 1 and 2 with zeros and apply the laws of indices.

David Quinn
  • 35,087
  • Okay that sounds fairly easy, I'm just a high school student so I'll have to look up the laws of indices haha :). Thanks for the reply! – Jack Curran Jan 20 '18 at 21:59
0

Hints:

  • the leading coefficient of the product of two polynomials is the product of the leading coefficients of those polynomials:

$$ (ax^m + \ldots [\text{powers of x < m}] \ldots)\cdot(bx^n + \ldots [\text{powers of x < n}] \ldots) \\ = ab\,x^{m+n} + \ldots [\text{powers of x < m+n}] \ldots $$

  • $\,f(x)\,$ is a product of monic polynomials i.e. polynomials with the leading coefficient $1$
dxiv
  • 77,867