Question

QUESTION 2 Factorize the following completely: \( 2.1 \quad x^{4}-49 \) \( 2.2 \quad 20-5 x^{2} \) \( 2.3 \quad x^{2}-8 x+15 \) \( 24-x^{3}-8^{3} \) \( 25 a^{2}\left(x^{2}+5 x-6\right) d t b^{2}\left(6-5 x-x^{2}\right) \)

Ask by Sherman Griffiths. in South Africa
Mar 14,2025

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Tutor-Verified Answer

Answer

Here are the complete factorizations for each expression: 1. \( x^{4} - 49 = (x^{2} - 7)(x^{2} + 7) \) 2. \( 20 - 5x^{2} = 5(2 - x)(2 + x) \) 3. \( x^{2} - 8x + 15 = (x - 5)(x - 3) \) 4. \( 24 - x^{3} - 8^{3} \) cannot be factored further. 5. \( 25a^{2}(x^{2} + 5x - 6)dtb^{2}(6 - 5x - x^{2}) = 25a^{2}dtb^{2}(-25)(x - 1)^{2}(x + 6)^{2} \)

Solution

Factor the expression by following steps: - step0: Factor: \(x^{2}-8x+15\) - step1: Rewrite the expression: \(x^{2}+\left(-3-5\right)x+15\) - step2: Calculate: \(x^{2}-3x-5x+15\) - step3: Rewrite the expression: \(x\times x-x\times 3-5x+5\times 3\) - step4: Factor the expression: \(x\left(x-3\right)-5\left(x-3\right)\) - step5: Factor the expression: \(\left(x-5\right)\left(x-3\right)\) Factor the expression \( 20-5 x^{2} \). Factor the expression by following steps: - step0: Factor: \(20-5x^{2}\) - step1: Factor the expression: \(5\left(4-x^{2}\right)\) - step2: Factor the expression: \(5\left(2-x\right)\left(2+x\right)\) Factor the expression \( x^{4}-49 \). Factor the expression by following steps: - step0: Factor: \(x^{4}-49\) - step1: Rewrite the expression: \(\left(x^{2}\right)^{2}-7^{2}\) - step2: Factor the expression: \(\left(x^{2}-7\right)\left(x^{2}+7\right)\) Factor the expression \( 25 a^{2}\left(x^{2}+5 x-6\right) d t b^{2}\left(6-5 x-x^{2}\right) \). Factor the expression by following steps: - step0: Factor: \(25a^{2}\left(x^{2}+5x-6\right)dtb^{2}\left(6-5x-x^{2}\right)\) - step1: Multiply the terms: \(25a^{2}dtb^{2}\left(x^{2}+5x-6\right)\left(6-5x-x^{2}\right)\) - step2: Multiply the terms: \(25a^{2}dtb^{2}\left(-\left(x^{2}+5x-6\right)^{2}\right)\) - step3: Factor the expression: \(25a^{2}dtb^{2}\left(-\left(x-1\right)^{2}\right)\left(x+6\right)^{2}\) - step4: Simplify the expression: \(-25a^{2}dtb^{2}\left(x-1\right)^{2}\left(x+6\right)^{2}\) Here are the complete factorizations for each of the expressions: ### 2.1 Factor \( x^{4}-49 \) This expression is a difference of squares: \[ x^{4}-49 = (x^{2}-7)(x^{2}+7) \] ### 2.2 Factor \( 20-5x^{2} \) We can factor out the common factor: \[ 20-5x^{2} = 5(4-x^{2}) = 5(2-x)(2+x) \] ### 2.3 Factor \( x^{2}-8x+15 \) This quadratic can be factored as: \[ x^{2}-8x+15 = (x-5)(x-3) \] ### 2.4 Factor \( 24-x^{3}-8^{3} \) This expression can be rewritten as: \[ 24 - x^{3} - 512 = 24 - (x^{3} + 512) \] However, it appears that no result could be found for this input. We can try to factor it using the sum of cubes formula, but it may require additional steps. ### 2.5 Factor \( 25a^{2}(x^{2}+5x-6)dtb^{2}(6-5x-x^{2}) \) This can be factored as: \[ 25a^{2}dtb^{2}(-25)(x-1)^{2}(x+6)^{2} \] Note that the second part \( (6-5x-x^{2}) \) can be rearranged and factored as well. If you need further assistance with the fourth expression or any other part, please let me know!

Answered by UpStudy AI and reviewed by a Professional Tutor

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Mind Expander

First off, when you see \( x^{4}-49 \), this is a difference of squares! You can factor it as \( (x^{2}-7)(x^{2}+7) \). Then there's \( 20 - 5x^{2} \), which you can pull out a common factor to get \( 5(4-x^{2}) \), and that further factors to \( 5(2-x)(2+x) \). For \( x^{2}-8x+15 \), this one’s a quadratic that factors nicely into \( (x-3)(x-5) \). Next, \( 24 - x^{3} - 8^{3} \) can be rearranged to \( 24 - (x+2)(x^{2}-2x+4) \). Lastly, for the more complex expression involving \( 25a^{2}(x^{2}+5x-6)dtb^{2}(6-5x-x^{2}) \), it factors down nicely into \( 25a^{2}(x-1)(x-6)dtb^{2}(-1)(x+5)(x-6) \). And voilà, your expression is completely factorized!

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