Question

How do you find the quotient of $(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48) \div (2y + 3)$?

Answer 1

Hint: Quotient is a quantity or expression resulting from the division of an expression by the other. In this question we have to divide $(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48)$ by $(2y + 3)$ and the resulting expression will be the quotient. For this we will use the long division method wherein we try to cut-off the terms of the dividend one by one. Also, we have,
$Dividend = (Quotient \times Divisor) + Remainder$

Complete step-by-step solution:
We have to divide $(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48)$ by $(2y + 3)$ and find the quotient.
We can write,
$(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48) = (Quotient \times (2y + 3)) + Remainder$
For this we will use the long division method.
We have to evaluate,
$(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48) \div (2y + 3)$
The degree of the expression of the quotient is the difference of the degrees of the dividend and the divisor. In this question the degree of the quotient will be $5 - 1 = 4$.
Let us assume the quotient is $(a{y^4} + b{y^3} + c{y^2} + dy + e)$, and the remainder is $r$.
We can write,
$(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48) = ((a{y^4} + b{y^3} + c{y^2} + dy + e) \times (2y + 3)) + r$
In long division method we try to cut-off the divisor term by term, i.e. we first try to cut-off the first term. We can see that $2y$ multiplied by $7{y^4}$ would give $14{y^5}$.
$(2y + 3) \times 7{y^4} = 14{y^5} + 21{y^4}$
We subtract $(14{y^5} + 21{y^4})$ from $(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48)$. We get,
$(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48) - (14{y^5} + 21{y^4}) = - 6{y^3} - 9{y^2} + 32y + 48$
So we get the first term of the quotient as $7{y^4}$ $(a = 7)$
We can see that the second term of the dividend with ${y^4}$ also got cut-off. This means that the next term of the quotient will also not be needed, i.e. for second term we can write $b = 0$.
Now we have to divide further using
$( - 6{y^3} - 9{y^2} + 32y + 48) = ((c{y^2} + dy + e) \times (2y + 3)) + r$
We can see that $2y$ multiplied by $ - 3{y^2}$ would give $ - 6{y^3}$.
$(2y + 3) \times - 3{y^2} = - 6{y^3} - 9{y^2}$
We subtract $( - 6{y^3} - 9{y^2})$ from $( - 6{y^3} - 9{y^2} + 32y + 48)$. We get,
$( - 6{y^3} - 9{y^2} + 32y + 48) - ( - 6{y^3} - 9{y^2}) = 32y + 48$
So we get the third term of the quotient as $ - 3{y^2}$ $(c = - 3)$
Again we see that the second term of the expression also gets cut-off. So the next term in quotient will also not be needed, i.e. for fourth term $d = 0$.
So now we have to divide further using,
$(32y + 48) = (e \times (2y + 3)) + r$
We can see that $2y$ multiplied by $16$ would give $32y$.
$(2y + 3) \times 16 = 32y + 48$
We subtract $(32y + 48)$ from $(32y + 48)$.
We get,
$(32y + 48) - (32y + 48) = 0$
So we get the fifth term of the quotient as $16$ $(e = 16)$.
We don’t have to take the division further.
The expression of the quotient becomes $(7{y^4} - 3{y^2} + 16)$, and the remainder is $0$.
We can write,
$(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48) = (7{y^4} - 3{y^2} + 16) \times (2y + 3)$
Hence, the quotient of $(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48) \div (2y + 3)$ is $(7{y^4} - 3{y^2} + 16)$.

Note: The degree of the quotient is equal to the difference between the degree of the dividend and the divisor. We can check the solution by equating RHS to LHS in $(14{y^5} + 21{y^4} - 6{y^3} - 9{y^2} + 32y + 48) = (7{y^4} - 3{y^2} + 16) \times (2y + 3)$