Answer: To find the quotient when 6x^4 – x³ - 26x² + 4x + 15 is divided by (x² - 4), we need to perform long division. Let's go through the steps:
Step 1: Write the dividend (6x^4 – x³ - 26x² + 4x + 15) and divisor (x² - 4).
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x² - 4 | 6x^4 - x³ - 26x² + 4x + 15
Step 2: Divide the first term of the dividend (6x^4) by the first term of the divisor (x²), which gives us 6x^2. Write this above the dividend.
6x^2
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x² - 4 | 6x^4 - x³ - 26x² + 4x + 15
Step 3: Multiply the divisor (x² - 4) by the quotient term (6x^2) and write the result below the dividend. Then, subtract this result from the dividend.
6x^2
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x² - 4 | 6x^4 - x³ - 26x² + 4x + 15
- (6x^4 - 24x²)
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22x² + 4x + 15
Step 4: Repeat the process with the new dividend (22x² + 4x + 15). Divide the first term of the new dividend (22x²) by the first term of the divisor (x²), which gives us 22. Write this above the new dividend.
6x^2 + 22
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x² - 4 | 6x^4 - x³ - 26x² + 4x + 15
- (6x^4 - 24x²)
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22x² + 4x + 15
- (22x² - 88)
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4x + 103
Step 5: Repeat the process with the new dividend (4x + 103). Divide the first term of the new dividend (4x) by the first term of the divisor (x²), which gives us 0. Write this above the new dividend.
6x^2 + 22 + 0
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x² - 4 | 6x^4 - x³ - 26x² + 4x + 15
- (6x^4 - 24x²)
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22x² + 4x + 15
- (22x² - 88)
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4x + 103
- (4x + 0)
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103
Step 6: There is no remainder, and the quotient is 6x^2 + 22 + 0 = 6x^2 + 22.
Now, to confirm that the remainder is 7 for the given equation 6x - x³ - 26x² + 4x + 8 = 0, substitute x = 1 (since the remainder is the value of the polynomial when the divisor is set to zero):
6(1) - (1)³ - 26(1)² + 4(1) + 8 = 6 - 1 - 26 + 4 + 8 = -9
The remainder is not 7, but rather -9. Therefore, there might be an error in the original equation or the value for the remainder.