Question

Find the value of x both the polynomial x²-x-6 and 3x²+8x+4 become zero?

Please answer with explanation​

Answer 1

Given polynomials :

• x² - x - 6
• 3x² + 8x + 4

To Find :

• Value of x for which the polynomials become zero.

Solution :

First figure out the value of x of the first polynomial.

We can do so by using the factorization method.

$\longrightarrow$ $\sf{x^2-x-6=0}$

$\longrightarrow$ $\sf{x^2-3x+2x-6=0}$

$\longrightarrow$ $\sf{x(x-3) +2(x-3) =0}$

$\longrightarrow$ $\sf{(x-3) (x+2)=0}$

$\longrightarrow$$\sf{x-3=0\:\:or\:\:x+2=0}$

$\longrightarrow$ $\sf{x=3\:\:or\:\:x=-2}$

We have two values of x.

$\bold{x=3\:\:or\:-2\:\:(1)}$

Now, moving to the next polynomial.

Figure out value of x using factorization method here too.

$\longrightarrow$ $\sf{3x^2\:+8x+4=0}$

$\longrightarrow$ $\sf{3x^2+6x+2x+4=0}$

$\longrightarrow$ $\sf{3x(x+2)+2(x+2)=0}$

$\longrightarrow$ $\sf{(x+2)\:\:(3x+2)=0}$

$\longrightarrow$ $\sf{x+2=0\:\:or\:\:3x+2=0}$

$\longrightarrow$ $\sf{x=-2\:\:or\:\:3x=-2}$

$\longrightarrow$ $\sf{x=-2\:\:or\:\:x=\dfrac{-2}{3}}$

Here, we have

$\bold{x=-2:\:or\:\dfrac{-2}{3}\:\:(2)}$

From equation (1) and (2), we clearly see x = -2 is the value for which both the polynomials become zero.