
How do you divide $\dfrac{{{x^2} - 2x - 15}}{{x + 3}}$ using polynomial long division?
Answer
452.1k+ views
Hint: Here are the steps required for dividing by a polynomial containing more than one term using the long division method.
• Make sure the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term.
• Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol.
• Multiply the answer obtained in the previous step by the polynomial in front of the division symbol.
• Subtract and bring down the next term.
• Repeat steps 2, 3, and 4 until there are no more terms to bring down.
• Write the final answer. The term remaining after the last subtract step is the remainder and must be written as a fraction in the final answer.
Complete step by step answer:
In this question, we have to find the quotient and the remainder of the division of\[{x^2} - 2x - 15\], the dividend, by $x + 3$, the divisor.
$x + 3\mathop {)\overline {{x^2} - 2x - 15} }\limits^{} $
The first step is that the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term. Here, the polynomial is already written in descending order and has not any missing term.
The second step is to divide the terms with the highest power inside the division symbol that is${x^2}$ by the term with the highest power outside the division symbol that is x. We will get the answer x.
$ \Rightarrow \dfrac{{{x^2}}}{x} = x$
The third step is to multiply the answer obtained in the previous step that is x by the polynomial in front of the division symbol isc$x + 3$. We will get the answer${x^2} + 3x$.
$ \Rightarrow x\left( {x + 3} \right) = {x^2} + 3x$
Subtract and bring down the next term. We will get the answer$ - 5x - 15$.
$x + 3\mathop {)\overline {{x^2} - 2x - 15} }\limits^x $
$\underline { \pm {x^2} \pm 3x} $
$ - 5x - 15$
Again we will repeat steps 2, 3, and 4.
The second step is to divide the terms with the highest power inside the division symbol that is -5x by the term with the highest power outside the division symbol that is x. We will get the answer -5.
$ \Rightarrow \dfrac{{ - 5x}}{x} = - 5$
The third step is to multiply the answer obtained in the previous step that is -5 by the polynomial in front of the division symbol is$x + 3$. We will get the answer$- 5x - 15$.
$ \Rightarrow - 5\left( {x + 3} \right) = - 5x - 15$
Subtract and bring down the next term. We will get the answer 0.
$x + 3\mathop {)\overline {{x^2} - 2x - 15} }\limits^{x - 5} $
$\underline { \pm {x^2} \pm 3x} $
$ - 5x - 15$
$\underline { \mp 5x \mp 15} $
$0$
Hence, the quotient is $x - 5$and the remainder is 0.
Note: We can verify our answer by the following formula.
$Dividend = (Divisor \times Quotient) + \operatorname{Re} mainder$
Here, dividend is${x^2} - 2x - 15$. Divisor is$x + 3$. Quotient is$x - 5$. Remainder is 0.
Let us take the right-hand side.
RHS=$(Divisor \times Quotient) + \operatorname{Re} mainder$
Let us put all the values.
$ \Rightarrow RHS = \left( {x - 5} \right) \times \left( {x + 3} \right) + 0$ $$
Simplify the above term.
$ \Rightarrow RHS = {x^2} + 3x - 5x - 15$
That is equal to
$ \Rightarrow RHS = {x^2} - 2x - 15$
Therefore,
$ \Rightarrow RHS = LHS$
• Make sure the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term.
• Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol.
• Multiply the answer obtained in the previous step by the polynomial in front of the division symbol.
• Subtract and bring down the next term.
• Repeat steps 2, 3, and 4 until there are no more terms to bring down.
• Write the final answer. The term remaining after the last subtract step is the remainder and must be written as a fraction in the final answer.
Complete step by step answer:
In this question, we have to find the quotient and the remainder of the division of\[{x^2} - 2x - 15\], the dividend, by $x + 3$, the divisor.
$x + 3\mathop {)\overline {{x^2} - 2x - 15} }\limits^{} $
The first step is that the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term. Here, the polynomial is already written in descending order and has not any missing term.
The second step is to divide the terms with the highest power inside the division symbol that is${x^2}$ by the term with the highest power outside the division symbol that is x. We will get the answer x.
$ \Rightarrow \dfrac{{{x^2}}}{x} = x$
The third step is to multiply the answer obtained in the previous step that is x by the polynomial in front of the division symbol isc$x + 3$. We will get the answer${x^2} + 3x$.
$ \Rightarrow x\left( {x + 3} \right) = {x^2} + 3x$
Subtract and bring down the next term. We will get the answer$ - 5x - 15$.
$x + 3\mathop {)\overline {{x^2} - 2x - 15} }\limits^x $
$\underline { \pm {x^2} \pm 3x} $
$ - 5x - 15$
Again we will repeat steps 2, 3, and 4.
The second step is to divide the terms with the highest power inside the division symbol that is -5x by the term with the highest power outside the division symbol that is x. We will get the answer -5.
$ \Rightarrow \dfrac{{ - 5x}}{x} = - 5$
The third step is to multiply the answer obtained in the previous step that is -5 by the polynomial in front of the division symbol is$x + 3$. We will get the answer$- 5x - 15$.
$ \Rightarrow - 5\left( {x + 3} \right) = - 5x - 15$
Subtract and bring down the next term. We will get the answer 0.
$x + 3\mathop {)\overline {{x^2} - 2x - 15} }\limits^{x - 5} $
$\underline { \pm {x^2} \pm 3x} $
$ - 5x - 15$
$\underline { \mp 5x \mp 15} $
$0$
Hence, the quotient is $x - 5$and the remainder is 0.
Note: We can verify our answer by the following formula.
$Dividend = (Divisor \times Quotient) + \operatorname{Re} mainder$
Here, dividend is${x^2} - 2x - 15$. Divisor is$x + 3$. Quotient is$x - 5$. Remainder is 0.
Let us take the right-hand side.
RHS=$(Divisor \times Quotient) + \operatorname{Re} mainder$
Let us put all the values.
$ \Rightarrow RHS = \left( {x - 5} \right) \times \left( {x + 3} \right) + 0$ $$
Simplify the above term.
$ \Rightarrow RHS = {x^2} + 3x - 5x - 15$
That is equal to
$ \Rightarrow RHS = {x^2} - 2x - 15$
Therefore,
$ \Rightarrow RHS = LHS$
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