What is the quotient of (6x^2-x-40) divided by (5+2x)
6x^2-x-40
= (2x+5)(3x-8)
so (6x^2-x-40) ÷ (5+2x) = 3x-8
To find the quotient of the polynomial (6x^2 - x - 40) divided by (5 + 2x), we can use polynomial long division. Here's how you can do it step by step:
Step 1: Arrange the terms of the dividend and the divisor in descending order of degree.
Dividend: 6x^2 - x - 40
Divisor: 5 + 2x
Step 2: Divide the highest-degree term of the dividend (6x^2) by the highest-degree term of the divisor (2x). The result is the first term of the quotient.
Quotient: 3x
Step 3: Multiply the divisor by the first term of the quotient (3x), and subtract the result from the dividend.
(6x^2 - x - 40) - (3x)(5 + 2x) = -7x - 40
Step 4: Bring down the next term from the dividend (-7x). This becomes the new dividend.
New dividend: -7x
Step 5: Divide the highest-degree term of the new dividend (-7x) by the highest-degree term of the divisor (2x). The result is the second term of the quotient.
New quotient: 3x - 4
Step 6: Multiply the divisor by the second term of the quotient (3x - 4), and subtract the result from the new dividend.
(-7x) - (3x - 4)(5 + 2x) = 12
Step 7: Since the new dividend (12) has a degree lower than the divisor, we have reached the end of the long division process.
Therefore, the quotient of (6x^2 - x - 40) divided by (5 + 2x) is 3x - 4, with a remainder of 12.
In polynomial form, this can be written as:
(6x^2 - x - 40)/(5 + 2x) = 3x - 4 + 12/(5 + 2x)