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The long  division:

Solve using synthetic division:

1  |  2  5  16  -23

   +  0  2   7    23

      -----------------

       2  7   23     0

The quotient is therefore 2x^2 + 7x + 23

The first number in the top line is 1 because x = 1 (x - 1 is the denominator). The 2, 5, 16, -23 are the coefficents of the numerator. You then use synthetic division to get the coefficients of the quotient and the remainder. You bring the first original coefficient (2) down untouched. You multiply that by the divisor (1) and add to the next coefficient (5). 5 + 2 = 7. You multiply that by the divisor and add to the next coefficient and so on. The last number you get is the remainder.

The pool is not realistic.

Hover the mouse over the ''log'' button.

This graph might help you answer your question

Given numbers? Equation? Which numbers and equation= ?

For me, I usually use factorization or quadratic formula. 

Factorization example:

\(x(x-5)=-4\\ x^2-5x+4=0\\ \color{red}(x-1)\color{aqua}(x-4)\color{black}=0\\ \text{Set each factor to 0.}\\ \color{red}x-1=0\\ \mathbf{\color{red}x=1}\\ \color{aqua}x-4=0\\ \mathbf{\color{aqua}x=4}\\ \therefore \color{red}x=1 \color{black} \text{ and } \color{aqua}x=4 \)

Quadratic formula example: 

\(x^2-x-1=0\\ x=\dfrac{-(-1)\pm\sqrt{(-1)^2-(4)(1)(-1)}}{2(1)}\\ \;\;=\dfrac{1\pm\sqrt5}{2}\\ \therefore x=\dfrac{1+\sqrt5}{2}\text{ and }x=\dfrac{1-\sqrt5}{2}\)

\(\text{Just for you to know: The solution of }ax^2+bx+c=0\\\text{ is }x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Complementation of the square example: (I usually call it completing the square)

\(x^2+8x+15=0\\ x^2+8x+16-1=0\\ (x+4)^2=1\\ x+4=\pm \sqrt1\\ \;\;\;\;\;\;\;\;\;\!=\pm 1\\ x= -3\text{ and }x=-5\)

\(\frac{1}{2}\times(6-14)-\frac{1}{3}\times(6-27)-3\\ =3-7-2+9-3\\ =0\)

\(\begin{array}{rlllll}&1&0&5&5&0\\-&&9&8&4&5\\-&-&-&-&-&-\\&&&7&0&5\end{array}\)

\(\begin{array}{rll}6(x+4)&=&3(x-2)\\2(x+4)&=&x-2\color{red}\text{(Divide by 3)}\\ 2x+8&=&x-2\color{red}\text{(Get rid of the brackets)}\\ x+8&=&-2\color{red}\text{(Minus x)}\\ x&=&-10\color{red}\text{(Minus 8)}\end{array} \)

Point P is 9 units from the center of a circle of radius 15.

How many different chords of the circle contain P and have integer lengths?

(15-9) * (15+9)  =   6 *  24  = 144

\(\begin{array}{|rcll|} \hline 144 &=& 2^4\cdot 3^2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 144 &=& (2\cdot 3) \cdot ( 2^3\cdot 3 ) \\ &=& 6 \cdot 24 \\\\ 144 &=& (2^3) \cdot ( 2\cdot 3^2 ) \\ &=& 8 \cdot 18 \\\\ 144 &=& (3^2) \cdot ( 2^4 ) \\ &=& 9 \cdot 16 \\\\ 144 &=& (2^2\cdot 3) \cdot (2^2\cdot 3) \\ &=& 12 \cdot 12 \\ \hline \end{array} \)

1. chord ( length = 30 = 6+24 ) 

2. chord ( length = 26 = 8+18 )

3. chord ( length = 26 = 18+8 )

4. chord ( length = 25 = 9+16 )

5. chord ( length = 25 = 16+9 )

6. chord ( length = 24 = 12+12 )

see chord theorem:

8d-3d-12=53

5d-12=53

5d=65

d=13. 

Done!! :D

Or am I really worrying so much?

Melody. My silly question.

b = 2A / h

What if h is 0 and the triangle is a straight line? b would be infinity for any values of A?

\(\frac{2}{3}+\frac{1}{2}\\ = \frac{4}{6}+\frac{3}{6}\\ = \frac{4+3}{6}\\ =\frac{7}{6}\\ =1\frac{1}{6}\)

Oops my mistake, 6 x 34 = 204.

y=23b+32023-204q

This representation on the right hand side of the equal sign could not be further simplified, So y is just equals 23b+32023-6×34q

WELL wrong one. Why would I have to reply to Guest #1 XD. I meant to make a new answer zone.

Why no one have asked me what is that cylindrical object with a lid? It's a void rubbish bin xDDDD

LOL good job Guest #2

Find it out yourself stupid fuckwit

Any number divided by 0 is undefined.

what plus what = 7 and what times what = -8 only using two numbers

\(\begin{array}{|rclcl|} \hline x^2+p \cdot x + q &=& 0 \qquad | \qquad -p=(x_1+x_2) & & q = x_1\cdot x_2 \\ x^2 -7 x -8 &=& 0 \qquad | \qquad -p=7 & & q =-8 \\\\ x_{1,2} &=& \frac{7\pm \sqrt{49-4\cdot(-8)} }{2} \\ x_{1,2} &=& \frac{7\pm \sqrt{81} }{2} \\ x_{1,2} &=& \frac{7\pm 9 }{2} \\\\ x_1 &=& \frac{7 + 9 }{2} \\ \mathbf{x_1} &\mathbf{=}& \mathbf{8} \\\\ x_2 &=& \frac{7 - 9 }{2} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{-1} \\ \hline \end{array}\)

8-1 = 7
8*(-1) = -8