Question

How do you multiply $(x-2)(x+3)?$

Answer 1

Hint: As we know that we have to multiply polynomial equations in two terms. Here we have to apply the bond mass rule in the above equation.
Distribution proper to apply on above term $a.(b+c)=a.b+a.c$

Complete step by step solution:
Here in the above equation we have to multiply to equation $(x-2)$ and $(x+3)$ . thus multiplying the above term \[(x-2)(x+3)\]
We can also write while multiplying we have
$x(x+3)-2(x+3)$
So, we had written as $x$ multiply with \[(x+3)\] and second multiplication is $-2$ multiply with $(x+3)$ when we are going to open and multiply the first bracket, while applying bodmas rule we have.
${{x}^{2}}+3x-2(x+3)$
Thus when opening the first bracket we obtained ${{x}^{2}}$ and $3x$ .
While further opening the second bracket we have. $2$ into $(x+3)$, we obtained ${{x}^{2}}+3x-2x-6$
When we are writing the similar term we have.
${{x}^{2}}+x(3-2)-6$
Here we get ${{x}^{2}}$ plus $x$ into $(3-2)$ and minus $6$ further solving $3$ minus $2$ we obtained $1$ but we have to give the greater number sign positive hence the value will be positive one. ${{x}^{2}}+x-6$
So, we get the value as ${{x}^{2}}$ plus $x$ minus $6$ we obtained while multiplying \[(x-2)\] and $(x+3)$

Hence we get ${{x}^{2}}+x-6$.

Note: Here we are going to multiply the polynomial we have to know about the term of the polynomial.
It will be easy when it has two polynomials. The level will increase as the polynomial goes up.