Solve 2x^90 - 4x^89 = 0: High-Degree Polynomial Equation

Polynomial Factoring with Common Factors

2x904x89=0 2x^{90}-4x^{89}=0

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find X
00:03 Factor with term 2X to the power of 89
00:13 Extract the common factor from parentheses
00:29 This is one solution that zeros the equation
00:39 Now let's check which solutions zero the second factor
00:45 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

2x904x89=0 2x^{90}-4x^{89}=0

2

Step-by-step solution

The equation in the problem is:

2x904x89=0 2x^{90}-4x^{89}=0 Let's pay attention to the left side:

The expression can be broken down into factors by taking out a common factor, The greatest common factor for the numbers and letters in this case is 2x89 2x^{89} since the power of 89 is the lowest power in the equation and therefore included both in the term where the power is 90 and in the term where the power is 89.

Any power higher than that is not included in the term where the power of 89 is the lowest, and therefore it is the term with the highest power that can be taken out of all the terms in the expression as a common factor for the variables.

For the numbers, note that the number 4 is a multiple of the number 2, so the number 2 is the greatest common factor for the numbers for the two terms in the expression.

Continuing and performing the factorization:

2x904x89=02x89(x2)=0 2x^{90}-4x^{89}=0 \\ \downarrow\\ 2x^{89}(x-2)=0 Let's continue and remember that on the left side of the equation that was obtained in the last step there is an algebraic expression and on the right side the number is 0.

Since the only way to get the result 0 from a product is for at least one of the factors in the product on the left side to be equal to zero,

Meaning:

2x89=0/:2x89=0/89x=0 2x^{89}=0 \hspace{8pt}\text{/}:2\\ x^{89}=0 \hspace{8pt}\text{/}\sqrt[89]{\hspace{6pt}}\\ \boxed{x=0}

Or:

x2=0x=2 x-2=0 \\ \boxed{x=2}

In summary:

2x904x89=02x89(x2)=02x89=0x=0x2=0x=2x=0,2 2x^{90}-4x^{89}=0 \\ \downarrow\\ 2x^{89}(x-2)=0 \\ \downarrow\\ 2x^{89}=0 \rightarrow\boxed{ x=0}\\ x-2=0\rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2} And therefore the correct answer is answer a.

3

Final Answer

x=0,2 x=0,2

Key Points to Remember

Essential concepts to master this topic
  • Factoring Rule: Find greatest common factor of all terms first
  • Technique: Extract 2x89 2x^{89} from both terms: 2x89(x2)=0 2x^{89}(x-2) = 0
  • Check: Substitute x=0 and x=2: 2(0)904(0)89=0 2(0)^{90} - 4(0)^{89} = 0

Common Mistakes

Avoid these frequent errors
  • Trying to solve without factoring first
    Don't attempt to solve 2x904x89=0 2x^{90} - 4x^{89} = 0 directly = impossible with such high powers! You'd need complex methods that don't work here. Always factor out the greatest common factor first to simplify the equation.

Practice Quiz

Test your knowledge with interactive questions

Break down the expression into basic terms:

\( 2x^2 \)

FAQ

Everything you need to know about this question

Why can't I just divide both sides by x to get rid of the variable?

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Never divide by a variable! If x = 0, you'd be dividing by zero, which is undefined. Plus, you'd lose the solution x = 0 completely. Always factor instead.

How do I find the greatest common factor with such high powers?

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Look for the lowest power of x in all terms. Here, x89 x^{89} is the lowest power, so it goes in the GCF along with the GCF of the coefficients (which is 2).

Why does the factored form 2x^89(x-2) = 0 give me two solutions?

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When a product equals zero, at least one factor must be zero. So either 2x89=0 2x^{89} = 0 (giving x = 0) or x2=0 x - 2 = 0 (giving x = 2).

Is x = 0 really a solution when it makes the power x^89?

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Yes! 089=0 0^{89} = 0 for any positive power. When you substitute x = 0, you get 2(0)4(0)=0 2(0) - 4(0) = 0 , which is true.

What if I missed factoring and tried other methods?

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Other methods like the quadratic formula won't work here because this isn't a quadratic equation. With such high degree polynomials, factoring is usually the only practical approach.

How can I check if x = 2 is correct?

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Substitute into the original equation: 2(2)904(2)89=2(2)8924(2)89=4(2)894(2)89=0 2(2)^{90} - 4(2)^{89} = 2(2)^{89} \cdot 2 - 4(2)^{89} = 4(2)^{89} - 4(2)^{89} = 0

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