Question

show that the polynomial 3x^3 +8x^2-1 has no integeral zeros
please someone answer step by step

Answer 1

Graph is made for justification.


Steps and Understanding :
1)
$f(x) = 3 {x}^{3} + 8 {x}^{2} - 1$
f'(x) = 9x^2 + 16x
f'(x) = 0
when
$9 {x}^{2} + 16x = 0 \\ = > x(9x + 16) = 0 \\ = > x = 0 \: or \: x \: = \frac{ - 16}{9} = - 1.778$
These are critical points where slope of graph changes its sign.

2) We observe that,
$f(0) = - 1 \\ f(1) = 10$
By Intermediate Value Theorem,

Since, f(0) < 0 , f(1) > 0.

there exists at least one root such that f(x) =0 which lies between x = 0 and x = 1 .
There is no integer between x = 0 and x = 1 , so at least one root is not integer.

3)
$f(0) = - 1 \\ f( - 1) = 4$
By Intermediate Value Theorem,

Since, f(0) < 0 , f(-1) > 0.

there exists at least one root such that f(x) =0 which lies between x = 0 and x = -1 .
There is no integer between x = 0 and x = -1 , so at least two roots which are not integer.

4) Since,
$f( - 2) > 0 \\ f( - 3) < 0$
By Intermediate Value Theorem,


there exists at least one root such that f(x) =0 which lies between x = -2 and x = -3
There is no integer between x = -2 and x = -3, so none roots are not integer.
As cubic equation has 3 roots possible.

And all roots are non - integral.

A graph is made for Justification.