How do you factor completely $16 x^{3} y^{4} + 8 x^{2} y^{3}$ $16x^3y^4+8x^2y^3$ ?

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How do you factor completely $16 x^{3} y^{4} + 8 x^{2} y^{3}$ $16x^3y^4+8x^2y^3$ ?

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Answer 1 · 2016-06-02T17:54:34

$8 x^{2} y^{3} (2 x y + 1)$ $8x^2y^3(2xy+1)$

Explanation:

$16 x^{3} y^{4} + 8 x^{2} y^{3}$ $16x^3y^4+8x^2y^3$

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

$16 x^{3} y^{4} = 2 \times 2 \times 2 \times 2 \times x \times x \times x \times y \times y \times y \times y$ $16x^3y^4=2xx2xx2xx2xx x xx x xx x xx yxxyxxyxxy$

$8 x^{2} y^{3} = 2 \times 2 \times 2 \times x \times x \times y \times y \times y$ $8x^2y^3=2xx2xx2xx x xx x xxyxxyxxy$

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

$16 x^{3} y^{4} = 2 \times 2 \times 2 \times 2 \times x \times x \times x \times y \times y \times y \times y$ $16x^3y^4=color(red)(2)xxcolor(red)(2)xxcolor(red)(2)xx2xx color(red)(x) xx color(red)(x) xx x xx color(red)(y)xxcolor(red)(y)xxcolor(red)(y)xxy$

$8 x^{2} y^{3} = 2 \times 2 \times 2 \times x \times x \times y \times y \times y$ $8x^2y^3=color(red)(2)xxcolor(red)(2)xxcolor(red)(2)xx color(red)(x) xx color(red)(x) xxcolor(red)(y)xxcolor(red)(y)xxcolor(red)(y)$

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

$= 2 \times 2 \times 2 \times x \times x \times y \times y \times y (2 \times x \times y + 1)$ $=color(red)(2)xxcolor(red)(2)xxcolor(red)(2)xx color(red)(x) xx color(red)(x) xxcolor(red)(y)xxcolor(red)(y)xxcolor(red)(y) ( color(blue)(2)xxcolor(blue)(x)xxcolor(blue)(y) + color(green)(1))$

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

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

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How do you factor completely $16 x^{3} y^{4} + 8 x^{2} y^{3}$ $16x^3y^4+8x^2y^3$ ?

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Explanation:

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How do you factor completely $16 x^{3} y^{4} + 8 x^{2} y^{3}$ $16x^3y^4+8x^2y^3$ ?