30 votes 30 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? $f(0)f(4)< 0$ $f(0)f(4)> 0$ $f(0)+f(4)> 0$ $f(0)+f(4)< 0$ Set Theory & Algebra gatecse-2014-set2 set-theory&algebra polynomials normal + – go_editor asked Sep 28, 2014 go_editor 5.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 97 votes 97 votes 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: $A$ Happy Mittal answered Sep 28, 2014 • edited Mar 25, 2021 by soujanyareddy13 Happy Mittal comment Share Follow See all 0 reply Please log in or register to add a comment.
20 votes 20 votes here we just need to draw possible graphs,no need to solve the whole equation and get values. shefali1 answered Oct 30, 2017 • edited Jun 8, 2018 by Milicevic3306 shefali1 comment Share Follow See all 3 Comments See all 3 3 Comments reply smsubham commented Mar 21, 2018 reply Follow Share 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) ) 0 votes 0 votes talha hashim commented Jan 12, 2019 reply Follow Share This should be the best solution because it cover both case when leading coefficient is positive or negative.nice @shifali 0 votes 0 votes pritishc commented Dec 22, 2019 reply Follow Share Isn't this the well known wavy curve method? 2 votes 2 votes Please log in or register to add a comment.
2 votes 2 votes 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. Regina Phalange answered Apr 3, 2017 Regina Phalange comment Share Follow See 1 comment See all 1 1 comment reply talha hashim commented Jul 4, 2018 reply Follow Share nice explanation @regina 0 votes 0 votes Please log in or register to add a comment.