Question

How do you factor ${x^4} + 2{x^3}y - 3{x^2}{y^2} - 4x{y^3} - {y^4}$?

Answer 1

Hint: This problem deals with factoring the given expression which is a polynomial of degree four. This polynomial is an expression with two variables \[x\] and \[y\]. We have to factorize the given expression in such a way that with its zeroes. So here first we factorize the four-degree polynomial into two degree polynomials and then factorize further.

Complete step-by-step solution:
The given expression ${x^4} + 2{x^3}y - 3{x^2}{y^2} - 4x{y^3} - {y^4}$ is equated to zero, as considered below:
$ \Rightarrow {x^4} + 2{x^3}y - 3{x^2}{y^2} - 4x{y^3} - {y^4} = 0$
First solving the equation with only $x$ terms as shown below, that is by considering the $y$ terms as 1.
$ \Rightarrow {x^4} + 2{x^3} - 3{x^2} - 4x - 1 = 0$
$ \Rightarrow \left( {{x^2} - x - 1} \right)\left( {{x^2} + 3x + 1} \right) = 0$
Here equating both the expressions to zero, as shown below to obtain the factors of $x$.
\[ \Rightarrow \left( {{x^2} - x - 1} \right) = 0\]
Now to find the roots of $x$ by using the formula of quadratic roots formula:
\[ \Rightarrow x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}\]
\[ \Rightarrow x = \dfrac{{1 \pm \sqrt {1 + 5} }}{2}\]
Now the second expression is equated as shown below:
\[ \Rightarrow \left( {{x^2} + 3x + 1} \right) = 0\]
$ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}$
On simplifying the expression as shown below:
$ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt 5 }}{2}$
Now factoring the expression ${x^4} + 2{x^3}y - 3{x^2}{y^2} - 4x{y^3} - {y^4}$ with the $y$ terms:
$ \Rightarrow {x^4} + 2{x^3}y - 3{x^2}{y^2} - 4x{y^3} - {y^4} = 0$
$ \Rightarrow \left( {{x^2} - xy - {y^2}} \right)\left( {{x^2} + 3xy + {y^2}} \right) = 0$
We know that the roots of $\left( {{x^2} - x - 1} \right) = 0$ are \[\dfrac{{1 \pm \sqrt {1 + 5} }}{2}\] , so the roots of the expression $\left( {{x^2} - xy - {y^2}} \right) = 0$ are equal to \[\dfrac{{1 \pm \sqrt {1 + 5} }}{2}y\].
Similarly the roots of the expression $\left( {{x^2} + 3xy + {y^2}} \right) = 0$ are equal to $\dfrac{{ - 3 \pm \sqrt 5 }}{2}y$.
$ \Rightarrow \left( {{x^2} - xy - {y^2}} \right)\left( {{x^2} + 3xy + {y^2}} \right) = 0$
$ \Rightarrow \left( {x - \dfrac{{1 + \sqrt {1 + 5} }}{2}y} \right)\left( {x - \dfrac{{1 - \sqrt {1 + 5} }}{2}y} \right)\left( {x - \dfrac{{ - 3 + \sqrt 5 }}{2}y} \right)\left( {x - \dfrac{{ - 3 - \sqrt 5 }}{2}y} \right) = 0$

Note: Please note that this problem deals with factoring the expression with the help of the roots of the quadratic equation formula. If the quadratic equation is expressed in the form of \[a{x^2} + bx + c = 0\], then the roots of the quadratic equation are given by the formula of the quadratic equation which is given by:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.