# Find the quotient & remainder for (2x^4+4x^2+2)/(x^2-1) using long division or synthetic division and find degree of quotient.

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Divide `2x^4+4x^2+2` by `x^2-1` using **synthetic division**:

The coefficients of the dividend are 2 0 4 0 2 and the coefficients of the divisor are 1 0 -1.

Drop the first coefficient of the divisor and negate the other coefficients to get 0 1. The division:

|2 0 4 0 2

1 | 2 0 6

0 | 0 0 0

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2 0 6 | 0 8

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**The solution is `2x^2+6+8/(2x^4+4x^2+2)` ; the degree of the quotient is 2.**

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The procedure:

(1)"drop" the first coefficient of the dividend. (the 2)

(2) Multiply the 2 along the diagonal by 0 and 1

(3) vertically add the second column resulting in 0

(4) multiply the 0 along the diagonal by 0 and 1

(5) vertically add the third column resulting in 6

(6) multiply the 6 along the diagonal by 0 and 1

(7) vertically add column 4 resulting in 0

(8) since further multiplication along the diagonal goes beyond the last column, vertically add all remaining columns.

(9) There are 2 numbers to the left of the bar -- thus the remainder has degree 2-1=1. Starting at the right the remainder will be 0x+8

(10) The quotient will be the remaining numbers, increasing in degree from the right or `2x^2+0x+6` . Since there were three numbers after the remainder, the degree of the quotient is 2.

The polynomial (2x^4+4x^2+2) has to be divided by (x^2-1) using long division.

x^2 - 1 | 2x^4 + 4x^2 + 2 | 2x^2 + 6

..............2x^4 - 2x^2

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........................6x^2 + 2

........................6x^2 - 6

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...................................8

**When (2x^4+4x^2+2) is divided by (x^2-1) the quotient is 2x^2 + 6 and the remainder is 8. The degree of the quotient is 2.**