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zepster313
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How to simplify factoring equations? 12 Nov 2024, 16:05
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I seem to get lost trying to factor equations. During the test calculators aren't allowed so using the quadratic formula seems a bit wasteful of time.
I have a doubt from the official quantitative guide that I wanted to clear

Problem. In this problem, how do they go from \(x(x-1)^2+5(x-1)^2\) to \((x+5)(x-1)^2\) in one step?

My work. Here I would try to do it the long way but I would get stuck on the last step with the red underline where I don't know how to factor a polynomial of the third degree.


Additionally, how can I quickly factor \(x^2-2x+1\) or any polynomial of the second degree without having to use the quadratic formula. I know this one is a bit easier but what about \(2x^2+6x-42\) for example? How would I quickly factor that to get both solutions of x?
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Bunuel
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How to simplify factoring equations? 13 Nov 2024, 01:08
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zepster313
I seem to get lost trying to factor equations. During the test calculators aren't allowed so using the quadratic formula seems a bit wasteful of time.
I have a doubt from the official quantitative guide that I wanted to clear

Problem. In this problem, how do they go from \(x(x-1)^2+5(x-1)^2\) to \((x+5)(x-1)^2\) in one step?

My work. Here I would try to do it the long way but I would get stuck on the last step with the red underline where I don't know how to factor a polynomial of the third degree.


Additionally, how can I quickly factor \(x^2-2x+1\) or any polynomial of the second degree without having to use the quadratic formula. I know this one is a bit easier but what about \(2x^2+6x-42\) for example? How would I quickly factor that to get both solutions of x?
Show more

1. How they go from \(x(x-1)^2 + 5(x-1)^2\) to \((x+5)(x-1)^2\):

In the expression \(x(x-1)^2 + 5(x-1)^2\), both terms contain \((x-1)^2\) as a common factor. So, you can factor out \((x-1)^2\) from each term:

\(x(x-1)^2 + 5(x-1)^2 = (x-1)^2 (x+5)\)

This is a straightforward factorization using the distributive property, where (x-1)^2 is factored out, leaving x + 5 in the parentheses.

2. How to quickly factor x^2 - 2x + 1

For \(x^2 - 2x + 1\), you can quickly identify it as a perfect square, specifically the square of a difference. This expression can be rewritten as:

\(x^2 - 2x + 1 = (x - 1)^2\)

Recognizing it as a perfect square saves time, as it follows the form \((a - b)^2 = a^2 - 2ab + b^2\), where here a = x and b = 1.

For more complex quadratics check these links:
Solving Quadratic Equations
Using Algebra vs. Logic on GMAT Quant Questions

­

7. Algebra



For more check Ultimate GMAT Quantitative Megathread

Hope it helps.­
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zepster313
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How to simplify factoring equations? 13 Nov 2024, 02:45
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Bunuel
zepster313
I seem to get lost trying to factor equations. During the test calculators aren't allowed so using the quadratic formula seems a bit wasteful of time.
I have a doubt from the official quantitative guide that I wanted to clear

Problem. In this problem, how do they go from \(x(x-1)^2+5(x-1)^2\) to \((x+5)(x-1)^2\) in one step?

My work. Here I would try to do it the long way but I would get stuck on the last step with the red underline where I don't know how to factor a polynomial of the third degree.


Additionally, how can I quickly factor \(x^2-2x+1\) or any polynomial of the second degree without having to use the quadratic formula. I know this one is a bit easier but what about \(2x^2+6x-42\) for example? How would I quickly factor that to get both solutions of x?

1. How they go from \(x(x-1)^2 + 5(x-1)^2\) to \((x+5)(x-1)^2\):

In the expression \(x(x-1)^2 + 5(x-1)^2\), both terms contain \((x-1)^2\) as a common factor. So, you can factor out \((x-1)^2\) from each term:

\(x(x-1)^2 + 5(x-1)^2 = (x-1)^2 (x+5)\)

This is a straightforward factorization using the distributive property, where (x-1)^2 is factored out, leaving x + 5 in the parentheses.
Show more
I still don't quite understand what you mean by factor out. Wouldn't removing the \((x-1)^2\) from each term make it just x+5 since you're removing the \((x-1)^2\) from both terms?
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Bunuel
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How to simplify factoring equations? 13 Nov 2024, 02:53
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zepster313
Bunuel
zepster313
I seem to get lost trying to factor equations. During the test calculators aren't allowed so using the quadratic formula seems a bit wasteful of time.
I have a doubt from the official quantitative guide that I wanted to clear

Problem. In this problem, how do they go from \(x(x-1)^2+5(x-1)^2\) to \((x+5)(x-1)^2\) in one step?

My work. Here I would try to do it the long way but I would get stuck on the last step with the red underline where I don't know how to factor a polynomial of the third degree.


Additionally, how can I quickly factor \(x^2-2x+1\) or any polynomial of the second degree without having to use the quadratic formula. I know this one is a bit easier but what about \(2x^2+6x-42\) for example? How would I quickly factor that to get both solutions of x?

1. How they go from \(x(x-1)^2 + 5(x-1)^2\) to \((x+5)(x-1)^2\):

In the expression \(x(x-1)^2 + 5(x-1)^2\), both terms contain \((x-1)^2\) as a common factor. So, you can factor out \((x-1)^2\) from each term:

\(x(x-1)^2 + 5(x-1)^2 = (x-1)^2 (x+5)\)

This is a straightforward factorization using the distributive property, where (x-1)^2 is factored out, leaving x + 5 in the parentheses.
I still don't quite understand what you mean by factor out. Wouldn't removing the \((x-1)^2\) from each term make it just x+5 since you're removing the \((x-1)^2\) from both terms?
Show more

When we say "factor out," we mean to treat (x-1)^2 as a common factor and place it outside the parentheses, just like you would with any other term that appears in multiple parts of an expression.

In this case:

\(x(x-1)^2 + 5(x-1)^2\)

becomes:

\((x-1)^2 * (x + 5)\)

This step does not "remove" (x-1)^2; instead, it pulls (x-1)^2 out as a common factor in front, leaving (x + 5) inside the parentheses. It’s like saying:

\(a * b + c * b = b * (a + c)\)

where b is the common factor. Here, (x-1)^2 is the common factor, and (x + 5) remains after factoring.

You might want to review some algebra fundamentals. Here’s a helpful link to get started with Algebra 101
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