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Factor the expression below.


x^(2) - 10x + 25

A. (x - 5)(x - 5)


B. (x + 5)(x + 5)


C. (x - 5)(x + 5)


D. 5(x2 - x + 5)

1 Answer

1 vote

Answer:

A. (x - 5)(x - 5)

Explanation:

We will do this the old fashioned way...just plain old factoring.

This polynomial is of the form


y=ax^2+bx+c

The product of a and c have to add up to equal the "middle" term, -10.

a = 1, b = -10, c = 25

a * c = 1 * 25 = 25

Now we need the factors of 25 to find the combination of factors that will result in a -10. The factors of 25 are: 1, 25 and 5, 5

5 and 5 add up to be 10, but since we need a -10, we will use -5 and -5. The product of -5 * -5 = 25, so we are not messing anything up by using the negative 5.

Putting them in order in standard form we have


x^2-5x-5x+25

Factor by grouping:


(x^2-5x)-(5x+25)

There is an x common to both terms in the first set of parenthesis, so we will factor that out; there is a 5 common to both terms in the second set of parenthesis, so we will factor that out:

x(x - 5) - 5(x - 5)

NOW what's common in both terms is the (x - 5) so we factor THAT out, and what's left gets grouped together:

(x - 5)(x - 5)

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