Problem 4 - Elementary and Intermediate Algebra

Question

a. Use multiplication to verify that the sum of two squares \(x^{2}+25\) does not factor as \((x+5)(x+5)\) b. Use multiplication to verify that the difference of two squares \(x^{2}-25\) factors as \((x+5)(x-5)\)

Step-by-Step Solution

Answer

Sum of squares doesn't factor as \((x+5)(x+5)\), but difference of squares factors as \((x+5)(x-5)\).

Step 1: Expand the suspect factorization

Expand the expression \((x+5)(x+5) = x^2 + 5x + 5x + 25 = x^2 + 10x + 25\).This is not equal to the original expression \(x^2 + 25\).

Step 2: Compare the expanded and original expression

Notice that \(x^2 + 10x + 25\) from the expanded expression includes a term \(10x\) which is missing from the original expression \(x^2 + 25\).Thus, \(x^2 + 25\) does not factor as \((x+5)(x+5)\).

Step 3: Expand the correct factorization

Expand the expression \((x+5)(x-5) = x^2 - 5x + 5x - 25 = x^2 - 25\).The result matches the original expression, \(x^2 - 25\).

Step 4: Verify the difference of squares identity

The expression \(x^2 - 25\) matches the result from the factorization of \((x+5)(x-5)\), confirming \(x^2 - a^2 = (x+a)(x-a)\).Thus, the difference of squares identity is verified.

Key Concepts

Sum of Squares

The sum of squares is an algebraic expression of the form \(a^2 + b^2\). Unlike the difference of squares, the sum cannot be factored into real linear factors. This means that expressions like \(x^2 + 25\) do not have a straightforward factorization in the real number system. It is important to remember that even though we might attempt the factorization as \((x+5)(x+5)\), this does not work because when expanded, it gives \(x^2 + 10x + 25\), which includes the term \(10x\) missing from the original expression.
Therefore, if you encounter a sum of squares, know that you usually cannot factor it into real-numbered binomials. Instead, it's stable in its given form under real numbers. If complex numbers are allowed, it might be factored, but that involves different algebra.
Thus, it is crucial to always verify factorization attempts by expanding and comparing terms.

Difference of Squares

The difference of squares is a special type of algebraic identity given by expressions like \(x^2 - 25\). Its general form is \(a^2 - b^2\), which can be neatly factored using the identity \((a+b)(a-b)\). This results in real, linear factors that perfectly match the original expression upon expansion.
For example, \(x^2 - 25\) factors into \((x+5)(x-5)\). When you expand \((x+5)(x-5)\), you get \(x^2 - 5x + 5x - 25 = x^2 - 25\). Notice, the middle terms \(-5x\) and \(+5x\) combine to zero, leaving us with \(x^2 - 25\).
This identity is very handy and widely used because it simplifies polynomial expressions into a product of binomials quickly and can be verified easily. Always remember to look for a pattern in \(a^2 - b^2\) when you have an expression that might be a difference of squares. This saves time and makes algebraic manipulation efficient.

Multiplication Verification

Verification through multiplication is an essential step in confirming the correctness of a factorization. It involves expanding the proposed factors to ensure they reconstruct the original expression. This was exemplified with the difference of squares \((x+5)(x-5)\) and the attempted sum of squares \((x+5)(x+5)\).
To verify, one would first expand the given factors. For example, \((x+5)(x-5)\) expands to \(x^2 - 5x + 5x - 25 = x^2 - 25\), matching perfectly with the original expression. However, the expansion of \((x+5)(x+5)\) results in \(x^2 + 10x + 25\), demonstrating it does not match \(x^2 + 25\).
This step is crucial as it eliminates mistakes and solidifies understanding. It teaches the importance of diligently checking your work in algebra to ensure accuracy. Being able to verify multiplications lets students confirm their factorization, avoiding errors in more complex problems.