Question

Find the zeroes of x2 + 2x - 48.​

Answer 1

Answer:

Here is your answer.

x²+2x-48

x²+(8-6)x-48

x²+8x-6x-48

x(x+8)-6(x+8)

(x+8)(x-6)

x=-8

x=6

So the zeros of the polynomial are -8 and 6

Hope it helps you.

Answer 2

$\large \underline{\underline{\textbf{Answer}}}$

$\large\boxed{\boxed{\mathtt{6 \: and\:-8}}}$

$\large \underline{\underline{\textbf{Step-by-step Explanation}}}$

Given :

$\mathsf{p(X) = {x}^2 + 2x - 48}$

We need to find the the zeroes of p(x).

Solution :

To find zeroes of any quadratic equation we use a formula known as Quadratic Formula that is

$\large \boxed{\boxed{\mathtt{\bigstar \: x = \dfrac{ - b \pm\sqrt{ {b}^{2} - 4ac} }{2a}}}}$

Now Here,

• a = coefficient of x^2 = 1
• b = coefficient of x = 2
• c = constant term = -48

*In order to use Quadratic Formula the Discriminate (D) of the equation should greater than or equal to zero.

Here, we have

$\mathsf{D = {b}^{2} - 4ac}$

Putting values of a , b , and c we have

$\longrightarrow \mathsf{D = {2}^{2} - 4(1)(-48)}$

$\longrightarrow \mathsf{D = 4 + 192}$

$\longrightarrow \mathsf{D = 196 > 0 }$

*Since D is greater than 0 we can use Quadratic Formula

Therefore we have

$\longrightarrow \mathtt{x = \dfrac{ - (2) \pm\sqrt{ {2}^{2} - 4(1)(-48)} }{2(1)}}$

$\longrightarrow \mathtt{x = \dfrac{ - (2) \pm\sqrt{196} }{2}}$

Therefore

$\longrightarrow \mathtt{x = \dfrac{-2 + 14}{2}\: or\: \dfrac{-2 - 14}{2}}$

That is --

$\longrightarrow \mathtt{x = 6\: or\: -8}$

Therefore zeroes of the polynomial are :

$\large\boxed{\mathtt{6 \: and\:-8}}$