Prove the condition below mentioned.

Question

Let $f(x)$, denote a polynomial in one variable with real coefficients, such that $f(a)=1$ for some real number a. Does there exist a polynomial $g(x)$ with real coefficients, such that, if $p(x)=f(x) g(x),$ then $p(a)=1$ $p^{\prime}(a)=0$ and $p^{\prime \prime}(a)=0 ?$ Justify your answer.

My approach: $p(x)=f(x) g(x),$

or,$p(a)=f(a) g(a)$

or,$p(a)= 1* g(a)$

Further I am getting no clue

Any hint will be highly appreciated

The implicit fact is that polynomials are always derived – David Morante Jul 11 '20 at 06:39 — David Morante, Jul 11 '20 at 06:39

score 2 · Accepted Answer · answered Jul 11 '20 at 06:36

We are going to deduce $g(x)$ from the conclusions:

$$p(a)=f(a)g(a)=1\Rightarrow g(a)=1$$ Also $$p'(a)=f'(a)g(a)+g'(a)f(a)$$ $$p'(a)=0=f'(a)+g'(a)\Rightarrow-f'(a)=g'(a)$$ Furthemore $$p''(a)=f''(a)g(a)+f'(a)g'(a)+g''(a)f(a)+g'(a)f'(a)=0$$ From the above then: $$p''(a)=0=f''(a)+g''(a)-2(f'(a))^2\Rightarrow g''(a)=2(f'(a))^2-f''(a)$$ Then :

$g (x)$ is such that $g(a)=1$ , $g'(a)=-f'(a)$ , $g''(a)=2(f'(a))^2-f''(a)$

Prove the condition below mentioned.

1 Answers1