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Ne4ueva [31]
3 years ago
9

Factor completely. 150m^2nz+20mn^2c-120m^2nc-25mn^2z

Mathematics
1 answer:
Anika [276]3 years ago
8 0

Answer:

The factor form is:

5mn\left(6m-n\right)\left(5z-4c\right)

Step-by-step explanation:

Here we have to find the factor of the expression:

150m^2nz+20mn^2c-120m^2nc-25mn^2z

So we need to take out the common terms, as we do for finding the greatest common factors.

Now the expression that is given can be re-written as:

150m^2nz+20mn^2c-120m^2nc-25mn^2z\\=150mmnz+20mnnc-120mmnc-25mnnz\\=30\cdot \:5nmmz+4\cdot \:5nmnc-24\cdot \:5nmmc-5\cdot \:5nmnz

Next, we will find the common terms, as follows;

30\cdot \:5nmmz+4\cdot \:5nmnc-24\cdot \:5nmmc-5\cdot \:5nmnz\\=5nm\left(30mz+4nc-24mc-5nz\right)\\

Now  we will factorise the expression:

\left(30mz+4nc-24mc-5nz\right)\\

as follows;

\left(30mz+4nc-24mc-5nz\right)\\=6m\left(5z-4c\right)+n\left(4c-5z\right)\\=\left(-4c+5z\right)\left(6m-n\right)\\

So the final factor form is:

150m^2nz+20mn^2c-120m^2nc-25mn^2z=5mn\left(6m-n\right)\left(5z-4c\right)

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Answer:

D) 0 = 2(x + 5)(x + 3)

Step-by-step explanation:

Which of the following quadratic equations has no solution?

We have to solve the Quadratic equation for all the options in other to get a positive value as a solution for x.

A) 0 = −2(x − 5)2 + 3

0 = -2(x - 5) × 5

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0 = -10x + 50

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x = 50/10

x = 5

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B) 0 = −2(x − 5)(x + 3)

Take each of the factors and equate them to zero

-2 = 0

= 0

x - 5 = 0

x = 5

x + 3 = 0

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Option B has a solution by one of its factors as a positive value of 5

C) 0 = 2(x − 5)2 + 3

0 = 2(x - 5) × 5

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0 = 10x -50

-10x = -50

x = -50/-10

x = 5

Option C has a solution of 5

D) 0 = 2(x + 5)(x + 3)

Take each of the factors and equate to zero

0 = 2

= 0

x + 5 = 0

x = -5

x + 3 = 0

x = -3

For option D, all the values of x are 0, or negative values of -5 and -3.

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