chase wants to factor x^2+10x+25 by grouping; however, Paige says it is a special product and can factor a different way. Using complete sentences, explain and demonstrate how both methods will result in the same factors.

## chase wants to factor x^2+10x+25 by grouping

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# 1 Answer

Chase and Paige are both correct. There are at least two different ways of factoring this expression.

Grouping: To factor this expression by grouping, you will need to rewrite it, by think of 10x as 5x + 5x. (why would you do that? No great reason, except if you noticed that 25 = 5 x 5, so we will try the factors of 25.)

Rewriting the expression, we would get

x

^{2}+ 10x + 25 = x^{2}+ (5x + 5x) + 25Now, since all of the individual terms are connected by addition, I can rearrange the parentheses to make it easier to think of the expression as two groups.

x

^{2}+ (5x + 5x) + 25 = (x^{2}+ 5x) + (5x + 25)Now we want to factor each of the groups in parentheses.

(x

^{2}+ 5x) = x(x+ 5) and (5x + 25) = 5(x + 5)So each group has one factor of (x + 5). That means we can use the distributive property in reverse, to combine the parts outside of the parentheses.

(x

^{2}+ 5x) + 5(x + 5) = x(x + 5) + 5(x + 5) = (x + 5)(x + 5)

And that is the factored form of the expression.

Special Product: In order to recognize this expression as a special product, you need to first recognize that 25 is a perfect square equal to 5 x 5. Since that is true, you might be tempted to think about the general form of any perfect square binomial.

That means, (x + a)

^{2}= x^{2}+ 2ax + a^{2}(trying multiplying it out with FOIL and that's what you will get.) This special form says that in a perfect square binomial the constant term of the expression will be a perfect square (25 = 5^{2}), and the middle term, (the 'x'-term) will be two times the separate parts of the binomial multiplied together ( in this case 2 times x times 5 = 10x.) The expression we wanted to factor looks exactly like that: x^{2}+ 10x + 25 = x^{2}+ two times x times 5 + 5^{2}.In both of these cases, you have to get used to imagining different ways of looking at the terms in the expression you are trying to factor. Patience and practice help a lot.

If you want a more visual explanation, let me know and I'll share a video with you.

Michael W

## Comments

ax2 + bx + c

Split b into two parts such that the product of the two parts equals the value of a * c

For example let's take a = 6, b = 19 and c = 15

a * c = 90

Split b = 19 into two parts which when multiplied yields a * c = 90. Now write the expression again as

6x2 + 9x + 10x + 15

This is equivalent to 3x(2x + 3) + 5(2x + 3). This can now be factored as (3x + 5) * (2x + 3)

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