
How do you find the quotient of $\left( 14{{x}^{2}}+7x \right)\div 7x$?
(a) Using linear formulas
(b) Using trigonometric identities
(c) Using algebraic properties
(d) None of these
Answer
536.4k+ views
Hint: To start with, we are given a problem where we have to find the value of the quotient from the given terms. We are here provided with the dividend $14{{x}^{2}}+7x$. From which we can take 7x common so that we can simplify and get the easier terms. Thus by cancelling out 7x, we get our needed solution to find our quotient.
Complete step-by-step answer:
According to the question, we are to find the quotient of $\left( 14{{x}^{2}}+7x \right)\div 7x$.
So, the term we are dividing is given as, $14{{x}^{2}}+7x$.
We can easily see, we can take 7x common from both the terms.
Now, by taking common 7x, we get, $14{{x}^{2}}+7x=7x\left( 2x+1 \right)$.
Let us divide this term by 7x. Thus, we are getting $\dfrac{7x\left( 2x+1 \right)}{7x}$.
Cancelling out 7x, we are getting 2x + 1.
Here we have used the formula, divisor $\times $ quotient + remainder = dividend.
So, the correct answer is “Option (a)”.
Note: This is always to be considered a easy problem to get the solution. But still we have to be cautious about the problem and the calculations given out there. Easier problems can get easily wrong with silly mistakes and give us wrong results. So, taking care of that is quite important.
Complete step-by-step answer:
According to the question, we are to find the quotient of $\left( 14{{x}^{2}}+7x \right)\div 7x$.
So, the term we are dividing is given as, $14{{x}^{2}}+7x$.
We can easily see, we can take 7x common from both the terms.
Now, by taking common 7x, we get, $14{{x}^{2}}+7x=7x\left( 2x+1 \right)$.
Let us divide this term by 7x. Thus, we are getting $\dfrac{7x\left( 2x+1 \right)}{7x}$.
Cancelling out 7x, we are getting 2x + 1.
Here we have used the formula, divisor $\times $ quotient + remainder = dividend.
So, the correct answer is “Option (a)”.
Note: This is always to be considered a easy problem to get the solution. But still we have to be cautious about the problem and the calculations given out there. Easier problems can get easily wrong with silly mistakes and give us wrong results. So, taking care of that is quite important.
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