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Question

Question: How do you factor \(64{x^4} + x{y^3}\) ?...

How do you factor 64x4+xy364{x^4} + x{y^3} ?

Explanation

Solution

To order to determine the factors of the above quadratic equation using the identity (A2B2)=(AB)(A+B)\left({{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right)
Formula:
(A3+B3)=(A+B)(A2A.B+B2)\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 64x4+xy364{x^4} + x{y^3},let it be f(x)f(x)
f(x)=64x4+xy3f(x) = 64{x^4} + x{y^3}
To simplify the above expression, pull out xx from both of the terms.
f(x)=(x)(64x3+y3)f(x) = (x)(64{x^3} + {y^3})
To find the factorization we’ll be writing the expression as
f(x)=(x)((4x)3+(y)3)f(x) = (x)\left( {{{(4x)}^3} + {{(y)}^3}} \right)
Consider 4x4xas A and yyas B and Applying Identity (A3+B3)=(A+B)(A2A.B+B2)\left( {{A^3} + {B^3}} \right) = \left( {A + B} \right)\left( {{A^2} - A.B + {B^2}} \right)
Now our equation becomes
f(x)=(x)(4x+y)(16x24xy+y2)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)(x) ,(4x+y)(4x + y),and(16x24xy+y2)(16{x^2} - 4xy + {y^2}).

Additional Information:
Quadratic Equation: A quadratic equation is a equation which can be represented in the form of ax2+bx+ca{x^2} + bx + cwhere xxis the unknown variable and a,b,c are the numbers known where a0a \ne 0.If a=0a = 0then 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=b24acD = {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=b+b24ac2ax1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} and x2=bb24ac2ax2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}
x1,x2 are root to quadratic equation ax2+bx+ca{x^2} + bx + c
Hence the factors will be (xx1)and(xx2)(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.