Suppose we have a generic polynomial of the form:
To know how many roots the polynomial can have, the first thing you should do is observe the term of greatest exponent.
For this case, the term of greatest exponent is 2.
Therefore, the polynomial has 2 roots.
Answer:
You must observe the term of the polynomial with greater exponent.
Answer:
(a) k'(0) = f'(0)g(0) + f(0)g'(0)
(b) m'(5) = 
Step-by-step explanation:
(a) Since k(x) is a function of two functions f(x) and g(x) [ k(x)=f(x)g(x) ], so for differentiating k(x) we need to use <u>product rule</u>,i.e., ![\frac{\mathrm{d} [f(x)\times g(x)]}{\mathrm{d} x}=\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20%5Bf%28x%29%5Ctimes%20g%28x%29%5D%7D%7B%5Cmathrm%7Bd%7D%20x%7D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20f%28x%29%7D%7B%5Cmathrm%7Bd%7D%20x%7D%5Ctimes%20g%28x%29%20%2B%20f%28x%29%5Ctimes%5Cfrac%7B%5Cmathrm%7Bd%7D%20g%28x%29%7D%7B%5Cmathrm%7Bd%7D%20x%7D)
this will give <em>k'(x)=f'(x)g(x) + f(x)g'(x)</em>
on substituting the value x=0, we will get the value of k'(0)
{for expressing the value in terms of numbers first we need to know the value of f(0), g(0), f'(0) and g'(0) in terms of numbers}{If f(0)=0 and g(0)=0, and f'(0) and g'(0) exists then k'(0)=0}
(b) m(x) is a function of two functions f(x) and g(x) [ 
]. Since m(x) has a function g(x) in the denominator so we need to use <u>division rule</u> to differentiate m(x). Division rule is as follows : 
this will give <em>
</em>
on substituting the value x=5, we will get the value of m'(5).
{for expressing the value in terms of numbers first we need to know the value of f(5), g(5), f'(5) and g'(5) in terms of numbers}
{NOTE : in m(x), g(x) ≠ 0 for all x in domain to make m(x) defined and even m'(x) }
{ NOTE : 
}
Answer:
It will be 57 points.
Step-by-step explanation:
