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3 years ago

3x(x-y)-p(y-x) factorise by taking out the common factor

Mathematics

1 answer:

puteri [66]3 years ago

4 0

Answer:

Rewrite as

3x(x-y) -p[-(x-y)]

We factored out a minus from the second bracket. I chose the second bracket arbitrarily... You can chose the 1st bracket if you want.

Now when those two minus interact... They became Positive(From rules of sign Multiplication)

3x(x-y) + p(x-y)

Now factor out (x-y) from both

(x-y) [ 3x + p]

So that's our answer!!

(x-y)[ 3x + p ].

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Answer:

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Step-by-step explanation:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}$

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$\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}$

$\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}$

$\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}$

Thus,

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3x(x-y)-p(y-x) factorise by taking out the common factor​

3x(x-y)-p(y-x) factorise by taking out the common factor