Q. Factor completely: \(9 – x^{2} = (3 – x)(3 + x)\).

Q: How do I factor 9 - x^2?

Use the difference of squares formula a^2 - b^2 = (a-b)(a+b) with a=3, b=x. So 9 - x^2 = (3 - x)(3 + x) (equivalently -(x-3)(x+3)).

Q: What if the expression is x^2 - 9 instead?.

Reorder or identify a=x, b=3. Then x^2 - 9 = (x - 3)(x + 3)..

Q: Can I factor a sum of squares like x^2 + 9?

Over the real numbers no; x^2 + 9 is irreducible in reals. Over complex numbers x^2 + 9 = (x + 3i)(x - 3i)..

Q: Should I factor out a negative sign first?

You can: 9 - x^2 = -(x^2 - 9) = -(x-3)(x+3). Either form is correct; choose the one that suits your context (leading coefficient, solving equations, etc.).

Q: What if one term isn't a perfect square, e.g., 8 - x^2?.

Over reals you can write 8 - x^2 = (\sqrt{8} - x)(\sqrt{8} + x), but that uses irrational coefficients. Over rationals it's not a nice factorization unless you factor out common factors first..

Q: How does factoring help solve 9 - x^2 = 0?.

Set each factor to zero: (3 - x)(3 + x) = 0 gives solutions x = 3 or x = -3.

Answer

\(9 − x^2\) is a difference of squares: \(3^2 − x^2 = (3 − x)(3 + x)\).

Final result: \(9-x^2=(3-x)(3+x)\)

Detailed Explanation

Problem

Factor completely: 9 − x²

Recognize the pattern.The expression 9 − x² is a difference of two squares because it can be written in the form a² − b².
Identify a and b.Write 9 as 3² and x² as (x)², so take
\( a = 3 \) and \( b = x \).
Apply the difference of squares formula.The identity for a difference of squares is
\( a^{2} – b^{2} = (a – b)(a + b) \).
Substituting \( a = 3 \) and \( b = x \) gives
\( 9 – x^{2} = (3 – x)(3 + x) \).
Check by expanding (optional verification).Multiply the factors to confirm:
\( (3 – x)(3 + x) = 3\cdot 3 + 3\cdot x – x\cdot 3 – x\cdot x = 9 – x^{2} \).
The expansion returns the original expression, so the factorization is correct.

Final answer: \( 9 – x^{2} = (3 – x)(3 + x) \)

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Algebra FAQs

How do I factor \(9 - x^2\)?

Use the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\) with \(a=3\), \(b=x\). So \(9 - x^2 = (3 - x)(3 + x)\) (equivalently \(-(x-3)(x+3)\)).

How do I recognize a difference of squares?

Both terms must be perfect squares and separated by subtraction. If an expression looks like \(-B\) with \(=a^2\), \(B=b^2\), it's a difference of squares: \(-B=(a-b)(a+b)\)..

What if the expression is \(x^2 - 9\) instead?.

Reorder or identify \(a=x\), \(b=3\). Then \(x^2 - 9 = (x - 3)(x + 3)\)..

Can I factor a sum of squares like \(x^2 + 9\)?

Over the real numbers no; \(x^2 + 9\) is irreducible in reals. Over complex numbers \(x^2 + 9 = (x + 3i)(x - 3i)\)..

Should I factor out a negative sign first?

You can: \(9 - x^2 = -(x^2 - 9) = -(x-3)(x+3)\). Either form is correct; choose the one that suits your context (leading coefficient, solving equations, etc.).

How do I check my factorization is correct?

What if one term isn't a perfect square, e.g., \(8 - x^2\)?.

Over reals you can write \(8 - x^2 = (\sqrt{8} - x)(\sqrt{8} + x)\), but that uses irrational coefficients. Over rationals it's not a nice factorization unless you factor out common factors first..

How does factoring help solve \(9 - x^2 = 0\)?.

Set each factor to zero: \((3 - x)(3 + x) = 0\) gives solutions \(x = 3\) or \(x = -3\).

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