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Quadratic Equations
MODULES
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SPEED CONCEPTS
PRACTICE
1) First term in the quotient will be the term which when multiplied with the highest-degree term in the divisor (i.e. x) will produce the highest-degree term in the dividend (5x3). 2) Write down the remainder after subtracting the product of the first term and the divisor from the dividend. |
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3) Second term in the quotient will be the term which when multiplied with the highest-degree term in the divisor (i.e. x) will produce the highest-degree term in the dividend and this process continues. 4) This process continues till the degree of the remainder is lower than that of the divisor. (When the divisor has a degree of 1, then the remainder will have a degree of 1 less than the divisor, which is 0. Therefore, the remainder will be a constant.) In the case to the right, when 5x3+6x2+3x−9 is divided by (x+2), we get a quotient of 5x2−4x+11 and a remainder of −31. |
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1) a will be the divisor. Write the down the coefficients of each of the terms in decreasing order of their powers. 2) Fill 0 under the first coefficient and add the two values and write it below. |
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3) Multiply the first sum (5) with the divisor (−2) and fill the product of (−10) under the next coefficient (6). Once again, add the two values (-4) and write it below. |
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4) Repeat the process outlined in step 3 till the end. The right-most number (−31) in the bottom is the remainder. The other numbers to the left of it are the coefficients of the quotient whose degree will be 1 less than the dividend. Note the the remainder in this case is −31. The other numbers in the bottom (i.e., 5,−4 and 11) form the coefficients of the quotient, i.e. 5x2−4x+11. |
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