Question

directions: factor each polynor

1. (x^2 - 36)

((x + 9)(x - 9))

1. (y^2 + 49)
2. (4y^2 - 81)
3. (m^2 - 64n^2)

(m - 8) (m + 8)

Explanation:

Problem 1: \(x^2 - 36\)

Step1: Identify the form

The expression \(x^2 - 36\) is a difference of squares, since \(x^2\) is a square (\((x)^2\)) and \(36\) is a square (\(6^2\)). The formula for factoring a difference of squares is \(a^2 - b^2=(a + b)(a - b)\).
Here, \(a = x\) and \(b = 6\) (because \(6^2=36\)).

Step2: Apply the formula

Substitute \(a=x\) and \(b = 6\) into the difference of squares formula:
\(x^2-36=(x + 6)(x - 6)\)
(Note: The original handwritten answer had a mistake, it should be \((x + 6)(x - 6)\) instead of \((x + 9)(x - 9)\))

Step1: Analyze the form

The expression \(y^2 + 49\) can be written as \(y^2+7^2\). In the real - number system, the sum of two squares \(a^2 + b^2\) (where \(a\) and \(b\) are real numbers) does not factor into linear factors with real coefficients. The formula for factoring a sum of squares in the complex - number system is \(a^2 + b^2=(a+bi)(a - bi)\), where \(i=\sqrt{- 1}\) is the imaginary unit.
Here, \(a = y\) and \(b = 7\).

Step2: Apply the complex - number factoring formula

Substitute \(a = y\) and \(b = 7\) into the formula \(a^2 + b^2=(a + bi)(a - bi)\):
\(y^2+49=(y + 7i)(y - 7i)\)
If we are restricted to the real - number system, the expression \(y^2 + 49\) is prime (cannot be factored into non - constant polynomials with real coefficients).

Step1: Identify the form

The expression \(4y^2-81\) is a difference of squares. We can rewrite \(4y^2=(2y)^2\) and \(81 = 9^2\). The formula for factoring a difference of squares is \(a^2 - b^2=(a + b)(a - b)\). Here, \(a = 2y\) and \(b=9\) (since \((2y)^2=4y^2\) and \(9^2 = 81\)).

Step2: Apply the formula

Substitute \(a = 2y\) and \(b = 9\) into the difference of squares formula:
\(4y^2-81=(2y + 9)(2y - 9)\)

Answer:

\((x + 6)(x - 6)\)

Problem 4: \(y^2+49\)