
How do you divide \[(3{x^2} - 29x + 56) \div (x - 7)\]?
Answer
524.1k+ views
Hint: Here we use the method of long division of polynomials where a polynomial is divided by another polynomial like normal division method.
The term that is being divided is called a dividend, the term that divides the term is called divisor, the value that we obtain on dividing is called quotient and the term that remains in the end is called remainder. We divide a by b giving us quotient q and remainder r, we write \[a = bq + r\]
* Long division method: when dividing \[a{x^n} + b{x^{n - 1}} + ....c\] by \[px + q\] we perform as
\[px + q)\overline {a{x^n} + b{x^{n - 1}} + ....c} ((a/p){x^{n - 1}} + ...\]
\[\underline { - a{x^n} + (qa/p){x^{n - 1}}} \]
\[0.{x^n} + (b - qa/p){x^{n - 1}}\]
Here we multiply the divisor with such a term that gives us the exact same term as the highest power in the dividend and then we proceed in the same way. We multiply the divisor with such a factor so we cancel out the highest power of the variable in it.
Complete step-by-step solution:
We are given the dividend as \[(3{x^2} - 29x + 56)\] and the divisor as \[(x - 7)\]
Now we use the long division method to divide the dividend by divisor.
\[x - 7)\overline {3{x^2} - 29x + 56} (3x - 8\]
\[\underline {3{x^2} - 21x} \]
\[ - 8x + 56\]
\[\underline { - 8x + 56} \]
\[0\]
When we divide a by b and we get q as quotient and r as remainder we write the equation
\[a = bq + r\]
Here we can write the equation as \[(3{x^2} - 29x + 56) = (3x - 8)(x - 7) + 0\]
\[\therefore \]The solution of \[(3{x^2} - 29x + 56) \div (x - 7)\] is \[(3x - 8)\]
Note: Alternate method:
We can form factors of the numerator using factorization method
We can write \[(3{x^2} - 29x + 56) \div (x - 7)\] as \[\dfrac{{(3{x^2} - 29x + 56)}}{{(x - 7)}}\]
We can write \[3{x^2} - 29x + 56 = 3{x^2} - 21x - 8x + 56\] because the product of 21 and 8 is 168 which is equal to product of 3 and 56.
Now we take common factors
\[ \Rightarrow 3{x^2} - 29x + 56 = 3x(x - 7) - 8(x - 7)\]
Collect the factors
\[ \Rightarrow 3{x^2} - 29x + 56 = (3x - 8)(x - 7)\]
Substitute this value in numerator
\[ \Rightarrow \dfrac{{(3{x^2} - 29x + 56)}}{{(x - 7)}} = \dfrac{{(3x - 8)(x - 7)}}{{(x - 7)}}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow \dfrac{{(3{x^2} - 29x + 56)}}{{(x - 7)}} = (3x - 8)\]
\[\therefore \]The solution of \[(3{x^2} - 29x + 56) \div (x - 7)\] is \[(3x - 8)\]
The term that is being divided is called a dividend, the term that divides the term is called divisor, the value that we obtain on dividing is called quotient and the term that remains in the end is called remainder. We divide a by b giving us quotient q and remainder r, we write \[a = bq + r\]
* Long division method: when dividing \[a{x^n} + b{x^{n - 1}} + ....c\] by \[px + q\] we perform as
\[px + q)\overline {a{x^n} + b{x^{n - 1}} + ....c} ((a/p){x^{n - 1}} + ...\]
\[\underline { - a{x^n} + (qa/p){x^{n - 1}}} \]
\[0.{x^n} + (b - qa/p){x^{n - 1}}\]
Here we multiply the divisor with such a term that gives us the exact same term as the highest power in the dividend and then we proceed in the same way. We multiply the divisor with such a factor so we cancel out the highest power of the variable in it.
Complete step-by-step solution:
We are given the dividend as \[(3{x^2} - 29x + 56)\] and the divisor as \[(x - 7)\]
Now we use the long division method to divide the dividend by divisor.
\[x - 7)\overline {3{x^2} - 29x + 56} (3x - 8\]
\[\underline {3{x^2} - 21x} \]
\[ - 8x + 56\]
\[\underline { - 8x + 56} \]
\[0\]
When we divide a by b and we get q as quotient and r as remainder we write the equation
\[a = bq + r\]
Here we can write the equation as \[(3{x^2} - 29x + 56) = (3x - 8)(x - 7) + 0\]
\[\therefore \]The solution of \[(3{x^2} - 29x + 56) \div (x - 7)\] is \[(3x - 8)\]
Note: Alternate method:
We can form factors of the numerator using factorization method
We can write \[(3{x^2} - 29x + 56) \div (x - 7)\] as \[\dfrac{{(3{x^2} - 29x + 56)}}{{(x - 7)}}\]
We can write \[3{x^2} - 29x + 56 = 3{x^2} - 21x - 8x + 56\] because the product of 21 and 8 is 168 which is equal to product of 3 and 56.
Now we take common factors
\[ \Rightarrow 3{x^2} - 29x + 56 = 3x(x - 7) - 8(x - 7)\]
Collect the factors
\[ \Rightarrow 3{x^2} - 29x + 56 = (3x - 8)(x - 7)\]
Substitute this value in numerator
\[ \Rightarrow \dfrac{{(3{x^2} - 29x + 56)}}{{(x - 7)}} = \dfrac{{(3x - 8)(x - 7)}}{{(x - 7)}}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow \dfrac{{(3{x^2} - 29x + 56)}}{{(x - 7)}} = (3x - 8)\]
\[\therefore \]The solution of \[(3{x^2} - 29x + 56) \div (x - 7)\] is \[(3x - 8)\]
Recently Updated Pages
How do you convert r6sec theta into Cartesian form class 10 maths CBSE

How do you solve dfrac5y3dfracy+72y6+1 and find any class 10 maths CBSE

If sin A+B1 and cos AB1 0circ le left A+B rightle 90circ class 10 maths CBSE

On the number line 10 is to the of zero class 10 maths CBSE

How do you solve 5xge 30 class 10 maths CBSE

In the following sentence supply a verb in agreement class 10 english CBSE

Trending doubts
Write an application to the principal requesting five class 10 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the median of the first 10 natural numbers class 10 maths CBSE

Write examples of herbivores carnivores and omnivo class 10 biology CBSE
