# f(x)=6xsq/ x-7 need to know the correct way too work this out. I got it wrong, not sure why

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If you want to calculate the division between 6x^2 (supposing that you mean by 6xsq = 6x square raised) and x-7, you could apply the rule of division with reminder.

Let's see how:

6x^2 = (x-7)*Q(x) + R(x)

Q(x) is the quotient and it is represented by a first degree polynomial:

Q(x) = ax+b

R(x) is the reminder and it is represented by a constant( a number). The degree of R(x) has to be smaller that the degree of x-7. Since the degree of the polynomial x-7 is 1, the degree of R(x) is 0.

Let's re-write the division:

6x^2 = (x-7)*(ax+b) + c

We'll remove the brackets from the right side:

6x^2 = ax^2 + bx - 7ax - 7b + c

We'll combine like terms from the right side:

6x^2 = ax^2 + x(b-7a) - 7b + c

The polynomial from the left side is equal with the polynomial from the right side, if and only if the correspondent coefficients are equal.

We'll re-write 6x^2 = 6x^2 + 0x + 0

**a = 6**

b - 7a = 0, but a = 6

b - 7*6 = 0

b - 42 = 0

We'll add 42 both sides:

**b = 42**

-7b + c = 0, but b = 42

c = 7b

c = 7*42

**c = 294**

**The result of the division is :**

** 6x^2/(x-7) = 6x + 42 + [294/(x-7)]**

f(x) = 6xsq/x-7.

Hope your kindself mean 6xsq to be 6 * x square = 6x^2.

If that is correct ,then the probem is to find 6x^2 /x-7 or 6x^2 divided by x-7. We do this by some rearrangement or long division method.

By rearrangement of numerator:

6x^2 = 6(x^2-7x+7x) = 6{(x^2-7x) +(7x-49)+49}

6x^2/(x-7) = 6{(x^2 - 7x) +7(x-7)+ 49}/(x-7) .

6x^2/(x-7) = 6{x(x-7) + 7(x-7) +49)}/(x-7) = 6{x+7(x-7)/(x-7) +49}/(x-7)

6x^2/(x-7) = 6{(x+7) + 49}/(x-y). Or

6x^2/(x-7) = 6x+294 is quotient (fully divided by x-7 ) and we get a remainder of 294 .

Long division method:

x-7) 6x^2 (** 6x +42** is the quotient.

6x^2 -42x

--------------

42x

42x- 294

-----------------------

**+294 **is the remainder

Hope this may help.