# 6x to the fourth -13x to the sceond -4 divided by 2x to the second-5 What is the answer?

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You need to perform the following division, hence, you may use reminder theorem, such that:

(16x^4 - 13x^2 - 4) = (2x^2 - 5)*q(x) + r(x)

Since the order of polynomial `2x^2 - 5` is 2, hence, the order of polynomial reminder is 1, such that:

`(16x^4 - 13x^2 - 4) = (2x^2 - 5)*(ax^2+bx+c) + (mx + n)`

Opening the brackets, yields:

`(16x^4 - 13x^2 - 4) = 2ax^4 + 2bx^3 + 2cx^2 - 5ax^2 - 5bx - 5c + mx + n`

You need to group the terms such that:

`(16x^4 - 13x^2 - 4) = 2ax^4 + 2bx^3 + x^2(2c - 5a) + x(-5b + m) - 5c + n`

Equating the coefficients of like powers yields:

`2a = 16 => a = 8`

`2b = 0 => b = 0`

`2c - 5a = -13 => 2c - 40 = -13 => 2c = 40 - 13 => c = 27/2`

`-5b + m = 0 => m = 0`

`- 5c + n = -4 => n = 5c - 4 => n = 135/2 - 4 => n = 127/2`

**Hence, performing the division `(16x^4 - 13x^2 - 4)/(2x^2 - 5)` yields the quotient `q(x) = 8x^2 + 27/2` and the reminder `r(x) = 127/2` .**

First, let's rewrite your question! (Always spell-check.)

6x to the fourth – 13x to the second -4 divided by 2x to the second -5

What is the answer??

6x⁴-13x²-4 ÷ 2x²-5

Now try solving this. Remember,

6x⁴ really is 6 times x² times x²