Question

Find the zeros, the sum of the zeros and the product of the zeros of the quadratic polynomial as follow:
$p\left(x\right)=3x^2-x-4$

Answer 1

In the given polynomial $3x^2-x-4$,

The first term is $3x^2$ and its coefficient is $3$.

The middle term is $-x$ and its coefficient is $-1$.

The last term is a constant term $-4$.

Multiply the coefficient of the first term by the constant $3\times -4=-12$.

We now find the factor of $12$ whose sum equals the coefficient of the middle term, which is $-1$ and then factorize the polynomial $3x^2-x-4$and equate it to $0$ as shown below:

$3x^2-x-4=03x^2+3x-4x-4=03x(x+1)-4(x+1)=0(x+1)(3x-4)=0(x+1)=0or(3x-4)=0x=-1,3x=4x=-1,x=\frac{4}{3}$

Therefore, the zeroes of the polynomial $3x^2-x-4$ are $-1,\frac{4}{3}$.

Now, the sum and the product of the zeroes of the given quadratic polynomial is as follows:

Sum of Zeroes$=-1+\frac{4}{3}=\frac{-3+4}{3}=\frac{1}{3}$

Product of Zeroes$=-1\times \frac{4}{3}=-\frac{4}{3}$

Hence, the sum of the zeroes is $\frac{1}{3}$ and the product of the zeroes is $-\frac{4}{3}$.