X3y2-zx2y-x4y3- - xy-z-x2y2

X^3y^2 zx^2y+x^4y^3 We can take x^2y as common from the given expression. We get nbsp; nbsp;x^2y(xy z+x^2y^2) So, x^3y^2 zx^2y+x^4y^3= x^2y× (xy z+x^2y^2) ...

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Byju's Answer

Standard VIII

Mathematics

Factorisation by Common Factors

x3y2-zx2y+x4y...

Question

$x^{3} y^{2} - z x^{2} y + x^{4} y^{3} =$ _______ $\times (x y - z + x^{2} y^{2})$

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Solution

$x^{3} y^{2} - z x^{2} y + x^{4} y^{3}$
We can take $x^{2} y$ as common from the given expression.
We get $x^{2} y (x y - z + x^{2} y^{2})$

So, $x^{3} y^{2} - z x^{2} y + x^{4} y^{3} =$ $x^{2} y \times (x y - z + x^{2} y^{2})$

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Similar questions

x^{3} y^{2} - z x^{2} y + x^{4} y^{3} =

_______

\times (x y - z + x^{2} y^{2})

x^{3} y^{2} - z x^{2} y + x^{4} y^{3} =

\times (x y - z + x^{2} y^{2})

Factorise: $x^{3} y^{2} - z x^{2} y + x^{4} y^{3}$

Question 6
If two positive integers a and b are written as $a = x^{3} y^{2} a n d b = x y^{2}$ where x, y are prime numbers, then HCF (a, b) is

A) $x y$
B) $x y^{2}$
C) $x^{3} y^{3}$
D) $x^{2} y^{2}$

x^{2} y^{2} - x y

for

x = 1

and

y = (- 1)

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x3y2−zx2y+x4y3= _______×(xy−z+x2y2)

x3y2−zx2y+x4y3 We can take x2y as common from the given expression. We get x2y(xy−z+x2y2) So, x3y2−zx2y+x4y3= x2y×(xy−z+x2y2)

$x^{3} y^{2} - z x^{2} y + x^{4} y^{3} =$ _______ $\times (x y - z + x^{2} y^{2})$

$x^{3} y^{2} - z x^{2} y + x^{4} y^{3}$
We can take $x^{2} y$ as common from the given expression.
We get $x^{2} y (x y - z + x^{2} y^{2})$

So, $x^{3} y^{2} - z x^{2} y + x^{4} y^{3} =$ $x^{2} y \times (x y - z + x^{2} y^{2})$