Question

Given f(x) = ax³ + 2x² + bx − 20 where x − 2 and x − 5 are two of its factors determine the real values of a and b

Answer 1

Given,

$f(x) = ax^3 + 2x^2 + bx − 20$

Factor x − 2 and x − 5

Now, substituting the value of x=2 and x =5 in the above equation

$ax^3+2x^2+bx-20=0$

x=2

$8a+8+2b-20=0$

Now,

$8a+2b=12$

$\Rightarrow 4a+b=6...(i)$

Similarly,

x=5

$125a+50+5b-20=0$

$\Rightarrow 125a+5b+30=0$

$\Rightarrow 25a+b=-6...(ii)$

From equation (i) and (ii)

$21a=-12$

$a=\frac{-4}{7}$

Now, substituting the value a in the above equation,

$b=6-\frac{16}{7}$

$\Rightarrow b= \frac{42-16}{7}$

$\Rightarrow b = \frac{26}{7}$