How do you factor $64{x^4} + x{y^3}$ ?
Answer
623.7k+ views
Hint:To order to determine the factors of the above quadratic equation using the identity $\left({{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right)$
Formula:
$\left( {{A^3} + {B^3}} \right) = \left( {A + B} \right)\left( {{A^2} - A.B + {B^2}} \right)$
Complete step by step solution:
Given a quadratic equation $64{x^4} + x{y^3}$,let it be $f(x)$
$f(x) = 64{x^4} + x{y^3}$
To simplify the above expression, pull out $x$ from both of the terms.
$f(x) = (x)(64{x^3} + {y^3})$
To find the factorization we’ll be writing the expression as
$f(x) = (x)\left( {{{(4x)}^3} + {{(y)}^3}} \right)$
Consider $4x$as A and $y$as B and Applying Identity $\left( {{A^3} + {B^3}} \right) = \left( {A + B}
\right)\left( {{A^2} - A.B + {B^2}} \right)$
Now our equation becomes
$f\left( x \right) = (x)(4x + y)(16{x^2} - 4xy + {y^2})$
Hence, we have successfully factorized our mathematical equation.
Therefore, the factors are$(x)$ ,$(4x + y)$,and$(16{x^2} - 4xy + {y^2})$.
Additional Information:
Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $a{x^2} + bx + c$where $x$is the unknown variable and a,b,c are the numbers known where $a \ne 0$.If $a = 0$then the equation will become a linear equation and will no longer be quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $D = {b^2} - 4ac$
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions
To find factors of Quadratic equation
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
x1,x2 are root to quadratic equation $a{x^2} + bx + c$
Hence the factors will be $(x - x1)\,and\,(x - x2)\,$.
Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.
Formula:
$\left( {{A^3} + {B^3}} \right) = \left( {A + B} \right)\left( {{A^2} - A.B + {B^2}} \right)$
Complete step by step solution:
Given a quadratic equation $64{x^4} + x{y^3}$,let it be $f(x)$
$f(x) = 64{x^4} + x{y^3}$
To simplify the above expression, pull out $x$ from both of the terms.
$f(x) = (x)(64{x^3} + {y^3})$
To find the factorization we’ll be writing the expression as
$f(x) = (x)\left( {{{(4x)}^3} + {{(y)}^3}} \right)$
Consider $4x$as A and $y$as B and Applying Identity $\left( {{A^3} + {B^3}} \right) = \left( {A + B}
\right)\left( {{A^2} - A.B + {B^2}} \right)$
Now our equation becomes
$f\left( x \right) = (x)(4x + y)(16{x^2} - 4xy + {y^2})$
Hence, we have successfully factorized our mathematical equation.
Therefore, the factors are$(x)$ ,$(4x + y)$,and$(16{x^2} - 4xy + {y^2})$.
Additional Information:
Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $a{x^2} + bx + c$where $x$is the unknown variable and a,b,c are the numbers known where $a \ne 0$.If $a = 0$then the equation will become a linear equation and will no longer be quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $D = {b^2} - 4ac$
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions
To find factors of Quadratic equation
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
x1,x2 are root to quadratic equation $a{x^2} + bx + c$
Hence the factors will be $(x - x1)\,and\,(x - x2)\,$.
Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.
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