How do you factor completely #16x^3y^4+8x^2y^3#?

Question

Savannah Anderson · Accepted Answer

#8x^2y^3(2xy+1)#

#16x^3y^4+8x^2y^3#

Let us write the factors of the terms given to us:

#16x^3y^4=2xx2xx2xx2xx x xx x xx x xx yxxyxxyxxy#

#8x^2y^3=2xx2xx2xx x xx x xxyxxyxxy#

Now, let us mark the common factors between the two terms:

#16x^3y^4=color(red)(2)xxcolor(red)(2)xxcolor(red)(2)xx2xx color(red)(x) xx color(red)(x) xx x xx color(red)(y)xxcolor(red)(y)xxcolor(red)(y)xxy#

#8x^2y^3=color(red)(2)xxcolor(red)(2)xxcolor(red)(2)xx color(red)(x) xx color(red)(x) xxcolor(red)(y)xxcolor(red)(y)xxcolor(red)(y)#

Write the common factors and put the remaining factors in brackets, with the appropriate sign:

#=color(red)(2)xxcolor(red)(2)xxcolor(red)(2)xx color(red)(x) xx color(red)(x) xxcolor(red)(y)xxcolor(red)(y)xxcolor(red)(y) ( color(blue)(2)xxcolor(blue)(x)xxcolor(blue)(y) + color(green)(1))#

#=color(red)(8x^2y^3)(color(blue)(2xy)+color(green)(1))#

Therefore, the factor of #16x^3y^4+8x^2y^3 = 8x^2y^3(2xy+1)#

Kennedy Adams · Answer

undefined To factor completely $16x^3y^4 + 8x^2y^3$, you can start by factoring out the greatest common factor, which is $8x^2y^3$. After factoring out the greatest common factor, the expression becomes $8x^2y^3(2xy + 1)$. Therefore, the completely factored form is $8x^2y^3(2xy + 1)$.