Factoring Challenge: Solve the Quadratic x²-6 as a Binomial Product

Difference of Squares with Irrational Numbers

x2−6=(x−—)⋅(x+—) x^2-6=(x-_—)\cdot(x+_—)

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Complete the missing value
00:04 We'll use the shortened multiplication formulas
00:16 We'll break down 6 into square root of 6 squared
00:22 A is X
00:25 B is square root of 6
00:28 B is the missing term
00:33 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

x2−6=(x−—)⋅(x+—) x^2-6=(x-_—)\cdot(x+_—)

2

Step-by-step solution

To solve the problem, we need to express x2−6x^2 - 6 in the form of (x−a)⋅(x+a)(x-a)\cdot(x+a) because this represents the difference of squares, which is expressed as (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2.

We are given x2−6x^2 - 6. Compare this to the formula x2−b2x^2 - b^2, it suggests that b2=6b^2 = 6.

The next step is to solve for bb by taking the square root of both sides:

b2=6⇒b=6b^2 = 6 \Rightarrow b = \sqrt{6}.

Thus, the missing number that completes the expression is 6\sqrt{6}.

Therefore, the solution to the problem is 6\sqrt{6}.

3

Final Answer

6 \sqrt{6}

Key Points to Remember

Essential concepts to master this topic
  • Pattern Recognition: x2−6 x^2 - 6 follows the difference of squares formula a2−b2 a^2 - b^2
  • Square Root Method: Set b2=6 b^2 = 6 , so b=6 b = \sqrt{6}
  • Verification: Check (x−6)(x+6)=x2−6 (x - \sqrt{6})(x + \sqrt{6}) = x^2 - 6 ✓

Common Mistakes

Avoid these frequent errors
  • Looking for rational factors instead of using square roots
    Don't try to factor x2−6 x^2 - 6 using integers like (x-2)(x+3) = wrong formula! This expression doesn't factor with rational numbers. Always recognize it as difference of squares: (x−6)(x+6) (x - \sqrt{6})(x + \sqrt{6}) .

Practice Quiz

Test your knowledge with interactive questions

Solve:

\( (2+x)(2-x)=0 \)

FAQ

Everything you need to know about this question

Why can't I factor this using regular integers like other quadratics?

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Because 6 is not a perfect square! When you have x2−6 x^2 - 6 , you need 6 \sqrt{6} which is irrational. Only expressions like x2−4 x^2 - 4 or x2−9 x^2 - 9 factor with integers.

How do I know this is a difference of squares pattern?

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Look for the form x2−number x^2 - \text{number} ! The difference of squares formula is a2−b2=(a−b)(a+b) a^2 - b^2 = (a-b)(a+b) . Here, a=x a = x and b2=6 b^2 = 6 .

What exactly is 6 \sqrt{6} ?

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6 \sqrt{6} is an irrational number approximately equal to 2.449. It's the exact value that, when squared, gives you 6. Don't try to simplify it further!

Can I check my answer by multiplying out?

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Absolutely! Use FOIL: (x−6)(x+6)=x2+x6−x6−(6)2=x2−6 (x - \sqrt{6})(x + \sqrt{6}) = x^2 + x\sqrt{6} - x\sqrt{6} - (\sqrt{6})^2 = x^2 - 6 . The middle terms cancel perfectly!

What if the question asked for x2+6 x^2 + 6 instead?

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Then it cannot be factored over real numbers! The sum of squares x2+6 x^2 + 6 has no real factors. Only difference of squares can be factored this way.

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