Posted by kingwendu on Saturday, December 17, 2011 at 8:26pm.
given that f(x)= 37x+5x^2x^3, factorise the polynomial completely, hence state the set of values of x for which the function is equal or less than zero .

function  Damon, Saturday, December 17, 2011 at 8:39pm
1 ( x^3  5 x^2 + 7 x  3)
I can see right away that x = 1 makes this zero because 88 = 0
divide polynomial by (x1)
******_x^2__4x____+_3_______________
x1 / x^3  5 x^2 + 7 x  3
*******x^3  1 x^2
*******
*********** 4 x^2 + 7 x 3
*********** 4 x^2 + 4 x
***********
*******************+ 3 x  3
*******************+ 3 x  3
remainder =0
so
1 (x1)(x^2  4 x + 3)
so
1 (x1)(x1)(x3)
zeros at 1, 1 (double,just hits axis) and 3
graph it remembering the  sign
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