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

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: $A$
edited

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

edited
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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) )
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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 https://en.wikipedia.org/wiki/... 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.

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nice explanation @regina

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