Question

How do you find the quotient of $(6{x^3} + 16{x^2} - 60x + 39) \div (2x + 10)$?

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 $(6{x^3} + 16{x^2} - 60x + 39)$ by $(2x + 10)$ and the resulting expression will be the quotient. For this we will use the long division method.
$Dividend = (Quotient \times Divisor) + Remainder$

Complete step by step solution:
We have given that,
\[{{x}^{4}}-5{{x}^{2}}+4=0\]
Taking out \[{{x}^{2}}\]as a common factor, we get
\[{{x}^{2}}\left( {{x}^{2}}-5 \right)+4=0\]
Substituting \[y={{x}^{2}}\] in the above equation, we get
\[y\left( y-5 \right)+4=0\]
Simplifying further we get
\[{{y}^{2}}-5y+4=0\]
Splitting the middle term, we obtained
\[{{y}^{2}}-4y-y+4=0\]
Taking out common factor by making pairs, we get
\[y\left( y-4 \right)-1\left( y-4 \right)=0\]
\[\left( y-1 \right)\left( y-4 \right)=0\]
Equating each common factor equal to 0, we get
\[y-1=0\ \ and\ \ y-4=0\]
Solving for the value of ‘y’, we get
\[y=1,4\]
Undo the substitution i.e. \[y={{x}^{2}}\],
\[{{x}^{2}}=1\ and\ {{x}^{2}}=4\]
Now solving
\[{{x}^{2}}=1\ \]
\[x=\sqrt{1\ }=\pm 1\]
Now solving
\[{{x}^{2}}=4\]
\[x=\sqrt{4\ }=\pm 2\]
\[x=\pm 2\]

Hence, the quotient of $(6{x^3} + 16{x^2} - 60x + 39) \div (2x + 10)$ is $(3{x^2} - 7x + 5)$.

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 $(6{x^3} + 16{x^2} - 60x + 39) = ((3{x^2} - 7x + 5) \times (2x + 10)) + ( - 11)$