
If $y-3$ is a factor of ${{y}^{3}}+2{{y}^{2}}-9y-18$, then find the other two factors.
Answer
626.4k+ views
Hint: Here one factor is given, we can find the other factors by polynomial division. First reduce the 3rd degree polynomial into a 2nd degree polynomial and then by reducing the 2nd degree polynomial we can find the other two factors.
Complete step-by-step solution -
Here, it is given that $y-3$ is a factor of the polynomial ${{y}^{3}}+2{{y}^{2}}-9y-18$. We have to find the other two factors.
First, we have to reduce the 3rd degree polynomial into a 2nd degree polynomial by polynomial division. The polynomial division is given by:
$y-3\overset{{{y}^{2}}+5y+6}{\overline{\left){\begin{align}
& {{y}^{3}}+2{{y}^{2}}-9y-18 \\
& {{y}^{3}}-3{{y}^{2}} \\
& \overline{0{{y}^{3}}\text{ +}5{{y}^{2}}-9y} \\
& \text{ }5{{y}^{2}}-15y \\
& \text{ }\overline{0{{y}^{2}}\text{ }+6y-18} \\
& \text{ }6y-18 \\
& \text{ }\overline{0y-0}\text{ } \\
\end{align}}\right.}}$
Therefore, by dividing the polynomial ${{y}^{3}}+2{{y}^{2}}-9y-18$ by $y-3$ we got the quotient as ${{y}^{2}}+5y+6$. i.e the 3rd degree polynomial is now reduced into a second degree polynomial .
Next, we can write the polynomial as:
${{y}^{3}}+2{{y}^{2}}-9y-18=(y-3)\left( {{y}^{2}}+5y+6 \right)$ ….. (1)
Now, we have to find the other two factors from the polynomial ${{y}^{2}}+5y+6$.
Consider ${{y}^{2}}+5y+6$ split $5y$as $2y+3y$. i.e we can write:
${{y}^{2}}+5y+6={{y}^{2}}+2y+3y+6$
For the above equation $y$ is common for the first two terms, so take $y$ outside and 3 is common for the last two terms, take 3 outside.
Hence, our equation becomes:
${{y}^{2}}+5y+6=y(y+2)+3(y+2)$
From the above equation since, $y+2$ is common, take it outside, we obtain:
${{y}^{2}}+5y+6=(y+2)(y+3)$
In the next step, we have to substitute the factors of the polynomial ${{y}^{2}}+5y+6$, $y+2$ and $y+3$ in equation (1). i.e. our equation (1) becomes:
${{y}^{3}}+2{{y}^{2}}-9y-18=(y-3)(y+2)(y+3)$
Therefore, the three factors of the polynomial ${{y}^{3}}+2{{y}^{2}}-9y-18$ are $y-3,y+2$ and $y+3$.
Note: Here in place of polynomial division you can also apply synthetic division to reduce the polynomial. If you know how to do synthetic division it is much simpler than polynomial division.
Complete step-by-step solution -
Here, it is given that $y-3$ is a factor of the polynomial ${{y}^{3}}+2{{y}^{2}}-9y-18$. We have to find the other two factors.
First, we have to reduce the 3rd degree polynomial into a 2nd degree polynomial by polynomial division. The polynomial division is given by:
$y-3\overset{{{y}^{2}}+5y+6}{\overline{\left){\begin{align}
& {{y}^{3}}+2{{y}^{2}}-9y-18 \\
& {{y}^{3}}-3{{y}^{2}} \\
& \overline{0{{y}^{3}}\text{ +}5{{y}^{2}}-9y} \\
& \text{ }5{{y}^{2}}-15y \\
& \text{ }\overline{0{{y}^{2}}\text{ }+6y-18} \\
& \text{ }6y-18 \\
& \text{ }\overline{0y-0}\text{ } \\
\end{align}}\right.}}$
Therefore, by dividing the polynomial ${{y}^{3}}+2{{y}^{2}}-9y-18$ by $y-3$ we got the quotient as ${{y}^{2}}+5y+6$. i.e the 3rd degree polynomial is now reduced into a second degree polynomial .
Next, we can write the polynomial as:
${{y}^{3}}+2{{y}^{2}}-9y-18=(y-3)\left( {{y}^{2}}+5y+6 \right)$ ….. (1)
Now, we have to find the other two factors from the polynomial ${{y}^{2}}+5y+6$.
Consider ${{y}^{2}}+5y+6$ split $5y$as $2y+3y$. i.e we can write:
${{y}^{2}}+5y+6={{y}^{2}}+2y+3y+6$
For the above equation $y$ is common for the first two terms, so take $y$ outside and 3 is common for the last two terms, take 3 outside.
Hence, our equation becomes:
${{y}^{2}}+5y+6=y(y+2)+3(y+2)$
From the above equation since, $y+2$ is common, take it outside, we obtain:
${{y}^{2}}+5y+6=(y+2)(y+3)$
In the next step, we have to substitute the factors of the polynomial ${{y}^{2}}+5y+6$, $y+2$ and $y+3$ in equation (1). i.e. our equation (1) becomes:
${{y}^{3}}+2{{y}^{2}}-9y-18=(y-3)(y+2)(y+3)$
Therefore, the three factors of the polynomial ${{y}^{3}}+2{{y}^{2}}-9y-18$ are $y-3,y+2$ and $y+3$.
Note: Here in place of polynomial division you can also apply synthetic division to reduce the polynomial. If you know how to do synthetic division it is much simpler than polynomial division.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

The number of corners in a cube are A 4 B 6 C 8 D class 8 maths CBSE

What are the methods of reducing friction. Explain

Advantages and disadvantages of science

Which part of nephron allows the selective reabsorption class 8 biology CBSE

