Simplify X 2 X 3

$\exponential{(x)}{two} - 2 10 - 3 $

Tick mark Image

Similar Problems from Web Search

a+b=-2 ab=one\left(-three\right)=-iii

Gene the expression past grouping. Start, the expression needs to be rewritten as x^{2}+ax+bx-3. To discover a and b, prepare upwardly a system to be solved.

a=-3 b=one

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the arrangement solution.

\left(x^{ii}-3x\right)+\left(x-3\right)

Rewrite x^{2}-2x-3 as \left(10^{2}-3x\right)+\left(x-three\correct).

x\left(x-iii\right)+x-3

Cistron out x in ten^{2}-3x.

\left(x-3\right)\left(x+1\right)

Gene out common term x-iii by using distributive belongings.

x^{two}-2x-3=0

Quadratic polynomial can be factored using the transformation ax^{two}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{ii} are the solutions of the quadratic equation ax^{2}+bx+c=0.

x=\frac{-\left(-2\correct)±\sqrt{\left(-two\correct)^{2}-iv\left(-three\correct)}}{2}

All equations of the form ax^{ii}+bx+c=0 tin be solved using the quadratic formula: \frac{-b±\sqrt{b^{ii}-4ac}}{2a}. The quadratic formula gives 2 solutions, one when ± is improver and 1 when it is subtraction.

10=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\correct)}}{2}

Square -2.

x=\frac{-\left(-2\right)±\sqrt{four+12}}{2}

Multiply -iv times -three.

x=\frac{-\left(-2\correct)±\sqrt{16}}{two}

Add together iv to 12.

ten=\frac{-\left(-2\correct)±iv}{2}

Have the foursquare root of xvi.

x=\frac{2±iv}{ii}

The reverse of -2 is 2.

10=\frac{half-dozen}{2}

Now solve the equation x=\frac{2±four}{ii} when ± is plus. Add two to 4.

10=\frac{-2}{2}

At present solve the equation x=\frac{2±iv}{2} when ± is minus. Subtract 4 from 2.

x^{2}-2x-3=\left(x-3\correct)\left(ten-\left(-1\right)\correct)

Factor the original expression using ax^{two}+bx+c=a\left(x-x_{one}\right)\left(x-x_{2}\right). Substitute iii for x_{1} and -1 for x_{ii}.

x^{ii}-2x-3=\left(ten-3\right)\left(x+1\right)

Simplify all the expressions of the form p-\left(-q\right) to p+q.

10 ^ 2 -2x -iii = 0

Quadratic equations such as this one tin exist solved by a new direct factoring method that does non require guess piece of work. To use the direct factoring method, the equation must be in the grade x^2+Bx+C=0.

r + due south = two rs = -iii

Allow r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(ten−s) where sum of factors (r+south)=−B and the product of factors rs = C

r = one - u s = one + u

Two numbers r and s sum upward to two exactly when the boilerplate of the ii numbers is \frac{1}{ii}*2 = 1. You tin can as well see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center past an unknown quantity u. Limited r and s with respect to variable u. <div way='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(1 - u) (1 + u) = -3

To solve for unknown quantity u, substitute these in the product equation rs = -3

ane - u^2 = -3

Simplify past expanding (a -b) (a + b) = a^2 – b^2

-u^ii = -three-one = -4

Simplify the expression past subtracting 1 on both sides

u^ii = iv u = \pm\sqrt{iv} = \pm ii

Simplify the expression by multiplying -one on both sides and take the square root to obtain the value of unknown variable u

r =one - 2 = -1 southward = 1 + 2 = three

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and south.