Question

show that the polynomial 3x^3+8x^2-1 has no integral zeros

Answer 1

Answer:

The proof is shown below

Step-by-step explanation:

Given polynomial is $3x^3+8x^2-1$

we have to show that above polynomial has no integral roots.

we use hit and trial method for finding the first root. Let it be $\frac{1}{3}$ which gives the remainder 0 by putting this value. Hence, $\frac{1}{3}$ is one of the root of above polynomial.

We can thus factorise the cubic as $(3x-1)(x^2+3x+1)$

Using the formula for finding the roots of the quadratic equation

$x=\frac{-3\pm\sqrt(9-4)}{2}$

$x=\frac{-3\pm\sqrt5}{2}$

which shows that the above polynomial have no integral roots

Hence, we have three distinct roots, none of which are integers.

Answer 2

Solution:

In a cubic equation there are no integer roots.

$3 x^{3}+8 x^{2}-1=0$

Integral zero theorem states that, -1 must be a factor of any integer works. Here we have two possibilities:

i) x = 1 and ii) x = -1

Let us now substitute the values in cubic equation.

i) $3 \times(1)^{3}+8 \times(1)^{2}-1$

$\Rightarrow 3+8-1$

$\Rightarrow 10 \neq 0$

ii) $3 \times(-1)^{3}+8 \times(-1)^{2}-1$

$\Rightarrow-3+8-1$

$\Rightarrow \mathbf{4} \neq \mathbf{0}$

Here integer roots are absent.

By repeating the same step again and again, we should find one root is given below:

$x=\frac{1}{3}$

Finally, factorize of cubic as,

$(3 x-1)\left(x^{2}+3 x+1\right)$

By formula, “quadratic value” will be

$x=-\frac{3}{2} \pm \frac{5}{\sqrt{2}}$

There is no integer in the 3 distinct roots.