5x 15 2x 24 4x

$\exponential{(ten)}{2} + 2 10 - 24 $

\left(x-4\right)\left(x+vi\right)

\left(10-iv\right)\left(x+half-dozen\right)

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a+b=2 ab=i\left(-24\correct)=-24

Factor the expression by group. Outset, the expression needs to be rewritten every bit x^{2}+ax+bx-24. To detect a and b, set upward a system to be solved.

-1,24 -ii,12 -3,8 -4,half-dozen

Since ab is negative, a and b have the reverse signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -24.

-1+24=23 -2+12=10 -three+eight=five -4+6=2

Summate the sum for each pair.

a=-4 b=vi

The solution is the pair that gives sum ii.

\left(10^{two}-4x\right)+\left(6x-24\right)

Rewrite x^{ii}+2x-24 as \left(x^{ii}-4x\right)+\left(6x-24\right).

x\left(x-4\correct)+six\left(ten-four\right)

Factor out 10 in the first and vi in the 2nd grouping.

\left(x-4\right)\left(x+6\right)

Factor out common term x-4 by using distributive property.

x^{2}+2x-24=0

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(10-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{two}+bx+c=0.

ten=\frac{-2±\sqrt{ii^{2}-iv\left(-24\right)}}{2}

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

x=\frac{-2±\sqrt{4-four\left(-24\right)}}{2}

Square 2.

10=\frac{-2±\sqrt{four+96}}{two}

Multiply -iv times -24.

x=\frac{-ii±\sqrt{100}}{2}

Add together 4 to 96.

x=\frac{-2±10}{two}

Take the foursquare root of 100.

x=\frac{8}{2}

Now solve the equation 10=\frac{-2±ten}{2} when ± is plus. Add together -2 to 10.

10=\frac{-12}{2}

Now solve the equation x=\frac{-2±10}{2} when ± is minus. Decrease 10 from -2.

x^{2}+2x-24=\left(x-4\right)\left(10-\left(-6\right)\right)

Factor the original expression using ax^{ii}+bx+c=a\left(x-x_{1}\right)\left(10-x_{2}\right). Substitute 4 for x_{i} and -6 for x_{two}.

ten^{2}+2x-24=\left(x-4\correct)\left(x+six\right)

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

x ^ ii +2x -24 = 0

Quadratic equations such equally this i can be solved by a new direct factoring method that does non require estimate work. To use the direct factoring method, the equation must be in the class x^two+Bx+C=0.

r + s = -ii rs = -24

Let r and s be the factors for the quadratic equation such that x^ii+Bx+C=(ten−r)(10−s) where sum of factors (r+southward)=−B and the production of factors rs = C

r = -ane - u s = -one + u

Ii numbers r and s sum upwardly to -2 exactly when the average of the two numbers is \frac{i}{2}*-2 = -1. You can also 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 by an unknown quantity u. Express r and s with respect to variable u. <div manner='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(-1 - u) (-1 + u) = -24

To solve for unknown quantity u, substitute these in the product equation rs = -24

1 - u^2 = -24

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

-u^2 = -24-1 = -25

Simplify the expression past subtracting 1 on both sides

u^2 = 25 u = \pm\sqrt{25} = \pm 5

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

r =-1 - 5 = -6 s = -1 + five = 4

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

5x 15 2x 24 4x,

Source: https://mathsolver.microsoft.com/en/solve-problem/%7B%20x%20%20%7D%5E%7B%202%20%20%7D%20%20%2B2x-24

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