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How do you solve \[3x{}^\text{2}-2x\]?

seo-qna
Last updated date: 14th Jun 2024
Total views: 372k
Views today: 5.72k
Answer
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Hint: We are given \[3x{}^\text{2}-2x\], we are asked to find the factor of this, to do so we will first understand the type of equation we have, once we get that we reduce the given equation into the standard \[\]form to simplify after that we will find the greatest common factor from each term then in the remaining term be factor using the middle term and lastly we compare it with zero and solve further
We use $a\times c$ in such a way that its sum or difference from the ‘b’ of the equation ax²+bx+c=0 We are given \[3x{}^\text{2}-2x\] we are asked to find the factor of it. Now to find the factor of the equation, we should see that as the highest power is $2$ so it is or $2$ degree polynomial. So, it is a quadratic equation.

Complete step by step answer:
As our Equation is quadratic equations and we know quadratic equations is given as \[ax{}^\text{2}+bx+c=0\] We reduce it to the standard form
We can see that our equation \[3x{}^\text{2}-2x\] is already in standard form
Now, to find its factor we will first find the possible greatest common factor of all these.
In $3$ and $2$ we can see that $1$ is the only possible term that can be separated, so we get – our equation stays like original as \[3x{}^\text{2}-2x\]
We can write this above equation as
\[3x{}^\text{2}-2x+0~\]\[\]
Now, we have middle term split on \[3x{}^\text{2}-2x+0~\]
We have \[a=3\text{ }b=-2\text{ }and\text{ }c=0~\]\[\]
So,
\[a\times c=3\times 0=0~\]
We have to find number whose sum or difference is \[-2\] and product is $0$ As product is zero one must be zero
Now we can see that \[0\times -2=0\text{ and }0-2=-2~\]
So, we use this to split the middle term.
So,
\[3x{}^\text{2}-2x+0=3x{}^\text{2}+\left( 0-2 \right)x+0~\]
Opening brackets, we get
\[~=3x{}^\text{2}+0x-2x+0~\]\[\]
We take common in the first $2$ terms and the last $2$ terms. So, we get – \[=x\left( 3x+0 \right)-x\left( -2+0 \right)\][ we can anything out of 0 as \[0=0\times x\]
As x is same, so we get –
\[=x\left( 3x+0-2+0 \right)~\]
Simplifying we get
 \[=x\left( 3x-2 \right)~\]
So, we get –
\[3x{}^\text{2}-2x+0=x\left( 3x-2 \right)~\]

Note: While find the middle term using factor of a c, we need to keep in mind that when the sign of ‘a’ and ‘c’ are same then ‘b’ is obtained by addition only, if the sign of ‘a’ and ‘c’ are different then ‘b’ can be obtained using only subtraction. Key point to remember is that the degree of the equation will also tell us about the number of solutions. Also $2$ degree means the given equation can have only $2$ Factor. We can easily find the factor of this. As we can see that our problem just has two terms so we will try to take the greatest possible common term out and then simply. We have \[3x{}^\text{2}-2x\]. We can see x is common to all. We take x from \[3x{}^\text{2}\] it will be left with 3x and we take x from \[-2x\] it will be left with \[-2\]. So \[3x{}^\text{2}-2x=x\left( 3x-2 \right).\]We get two factors. It cannot be factored more. This is our factor form \[3x{}^\text{2}-2x=x\left( 3x-2 \right).\]