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+17 votes

A non-zero polynomial $f(x)$ of degree 3 has roots at $x=1$, $x=2$ and $x=3$. Which one of the following must be TRUE? 

  1. $f(0)f(4)< 0$
  2. $f(0)f(4)> 0$
  3. $f(0)+f(4)> 0$
  4. $f(0)+f(4)< 0$
asked in Set Theory & Algebra by Veteran (96.1k points) | 1.5k views

3 Answers

+57 votes
Best answer
The roots are $x=1, x=2$, and $x=3.$

So polynomial is $f(x) = (x-1)(x-2)(x-3)$

$f(0) = -6, f(4) = 6$

So $f(0)f(4) < 0$.

Correct Answer: $A$
answered by Boss (11.4k points)
edited by
+10 votes

here we just need to draw possible graphs,no need to solve the whole equation and get values.

answered by Active (1.6k points)
edited by
Why draw graph when you can solve it directly? f(0) value is there already, f(4) can be calculated as we have f(x). ( f is of degree 3 all roots are there so we can get f(x) )
This should be the best solution because it cover both case when leading coefficient is positive or negative.nice @shifali
+1 vote

The graph of a degree 3 polynomial f(x) = a0 + a1x + a2(x^2) + a3(x^3), where a3 ≠ 0 is a cubic curve, as can be seen here Now as given, the polynomial is zero at x = 1, x = 2 and x = 3, i.e. these are the only 3 real roots of this polynomial. Hence we can write the polynomial as f(x) = K (x-1)(x-2)(x-3) where K is some constant coefficient. Now f(0) = -6K and f(4) = 6K ( by putting x = 0 and x = 4 in the above polynomial ) and f(0)*f(4) = -36(k^2), which is always negative. Hence option A. We can also get the answer by just looking at the graph. At x < 1, the cubic graph (or say f(x) ) is at one side of x-axis, and at x > 3 it should be at other side of x-axis. Hence +ve and -ve values, whose multiplication gives negative.

answered by Loyal (9.3k points)
nice explanation @regina

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