Question

One root of the equation$\left(x+1\right)\left(x+3\right)\left(x+2\right)\left(x+4\right)=120$is

• A. $-1$
• B. $2$
• C. $0$
• D. $1$

Answer 1

The correct option is D

$1$

Zeros of a polynomial can be defined as the points where the polynomial becomes zero as a whole. A polynomial having a value of zero$\left(0\right)$is called zero polynomial.

Given$\left(x+1\right)\left(x+3\right)\left(x+2\right)\left(x+4\right)=120$

Let's substitute the values from the option, in the equation and check.

(A) Put$x=-1$in given equation.

$$\begin{gathered} =\left(-1+1\right)\left(-1+3\right)\left(-1+2\right)\left(-1+4\right) \\ =0\times 2\times 1\times 3 \\ =0\ne 120 \end{gathered}$$

(B) Put$x=2$in given equation.

$$\begin{gathered} =\left(2+1\right)\left(2+3\right)\left(2+2\right)\left(2+4\right) \\ =3\times 5\times 4\times 6 \\ =360\ne 120 \end{gathered}$$

(C) Put$x=0$in given equation.

$$\begin{gathered} =\left(0+1\right)\left(0+3\right)\left(0+2\right)\left(0+4\right) \\ =1\times 3\times 2\times 4 \\ =24\ne 120 \end{gathered}$$

(D) Put$x=1$in given equation.

$$\begin{gathered} =\left(1+1\right)\left(1+3\right)\left(1+2\right)\left(1+4\right) \\ =2\times 4\times 3\times 5 \\ =120=120 \end{gathered}$$

Since, $x=1$satisfying the equation, therefore the root of the equation is$x=1$.