Question

f(x) is a monic quadratic polynomial with f(-1)=0 and f(1)+f(2)=0 .FIND f(8/5)=


PLEASE EXPLAIN IT

Answer 1

If f(x) is a monic quadratic polynomial with f(-1) = 0 and f(1) + f(2) = 0, then f(8/5) = 0

Step-by-step explanation:

Let

$f(x)=ax^2+bx+c$  a ≠ 0

Given

$f(-1)=0$

$\implies a(-1)^2+b(-1)+c=0$

$\implies a-b+c=0$

$\implies 1-\frac{b}{a}+\frac{c}{a}=0$

Let b/a = u and c/a = v

Then,

$-u+v=-1$   ........... (1)

Also,

$f(1)+f(2)=0$

$\implies (a+b+c)+(4a+2b+c)=0$

$\implies 5a+3b+2c=0$

$\implies 3\frac{b}{a}+2\frac{c}{a}=-5$

$\implies 3u+2v=-5$ .............. (2)

Multiplying eq (1) by 3 and adding it in eq (2)

$3v+2v=-3-5$

$\implies 5v=-8$

$\implies v=-\frac{8}{5}$

Thus,

$u=1+v=1-\frac{8}{5}$

$\implies u=\frac{-3}{5}$

Thus,

$\frac{b}{a}=-\frac{3}{5}$

$\frac{c}{a}=-\frac{8}{5}$

Therefore, the polynomial is

$f(x)=x^2-\frac{3}{5}x-\frac{8}{5}$

or, $f(x)=5x^2-3x-8$

Therefore,

$f(\frac{8}{5})=5(\frac{8}{5})^2-3\times\frac{8}{5}-8$

$\implies f(\frac{8}{5})=\frac{64}{5}-\frac{24}{5}-8$

$\implies f(\frac{8}{5})=\frac{40}{5}-8$

$\implies f(\frac{8}{5})=8-8=0$

Hope this answer is helpful.

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Answer 2

Given :  f(x) is a monic quadratic polynomial with f(-1)=0 and f(1)+f(2)=0

To find :   f(8/5)=

Solution:

Lets take another Simpler  approach in addition to solution above  by sonuvuce

f(-1) = 0

=> -1 is a zero

=> x + 1  is a factor

Let say other zero = a

=> (x - a) is a factor

f(x) = (x + 1)(x - a)

f(1)  + f(2) = 0

=> (1 + 1)(1 - a)  + (2 + 1)(2 - a)  = 0

=> 2 - 2a  + 6 - 3a = 0

=> a  = 8/5

8/5 is other zero

Hence f(8/5)  = 0

additional info :

f(x) = (x + 1)(x - 8/5)    =  x²  -3x/5 - 8/5  = 5x² - 3x  - 8

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