1

A polynomial $f(x)=x^3+2x^2+3x+4$ has roots $a, b, c$. Find the value of $(a^2-1)(b^2-1)(c^2-1)$.

I found this question in a class test. The solution my teacher gave after the test was $-20$. I am not sure of how the answer can be derived. Please help.

Teddy38
  • 3,309
Nishchay Saligram
  • 29
  • 7
    Look at $f(x)f(-x)$ evaluated at $x=1$. If $f(x)=(x-a)(x-b)(x-c)$, then $f(x)f(-x)=(-x^2+a^2)(-x^2+b^2)(-x^2+c^2)$. – plop Jun 04 '21 at 14:22
  • Hi Nishchay. Can you please tell us about the things you have tried? Click "Edit" above. If you haven't tried anything, this question perhaps doesn't belong here. Also we suggest you review this material: https://math.meta.stackexchange.com/q/9959 – 311411 Jun 04 '21 at 14:29

2 Answers2

2

Let $$f(x)={t_1}x^3+{t_2}x^2+{t_3}x+{t_4}$$Using Vieta's theorem, we can know $$a+b+c=-\frac{t_2}{t_1}=-2$$$$ab+bc+ca=\frac{t_3}{t_1}=3$$$$abc=-\frac{t_4}{t_1}=-4$$Then we expansion the formula and put those value into the formula.

Waldeinsamkeit
  • 21
2

To expand on @plops comment

you can use the fact

$$f(x) = k(a-x)(b-x)(c-x)$$

Now we can recognize the conjugate rule in every factor of the product $$(a^2-1) = (a-1)(a+1)\\...$$ So the expression is simply $$\frac{f(1)f(-1)}{k^2}$$

Now to find $k$ investigate the $x^3$ term.

What remains is simple arithmetics. You can probably finish it.

mathreadler
  • 25,824