Type a math problem

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Type a math problem

Factor

2\left(x+2\right)\left(x+6\right)

$2(x+2)(x+6)$

Solution Steps

Steps Using the Quadratic Formula

Steps Using Direct Factoring Method

Solution Steps

2 x ^ { 2 } + 16 x + 24

$2x_{2}+16x+24$

Factor out 2.

Factor out $2$.

2\left(x^{2}+8x+12\right)

$2(x_{2}+8x+12)$

Consider x^{2}+8x+12. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

Consider $x_{2}+8x+12$. Factor the expression by grouping. First, the expression needs to be rewritten as $x_{2}+ax+bx+12$. To find $a$ and $b$, set up a system to be solved.

a+b=8 ab=1\times 12=12

$a+b=8$ $ab=1×12=12$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $12$.

1,12 2,6 3,4

$1,12$ $2,6$ $3,4$

Calculate the sum for each pair.

Calculate the sum for each pair.

1+12=13 2+6=8 3+4=7

$1+12=13$ $2+6=8$ $3+4=7$

The solution is the pair that gives sum 8.

The solution is the pair that gives sum $8$.

a=2 b=6

$a=2$ $b=6$

Rewrite x^{2}+8x+12 as \left(x^{2}+2x\right)+\left(6x+12\right).

Rewrite $x_{2}+8x+12$ as $(x_{2}+2x)+(6x+12)$.

\left(x^{2}+2x\right)+\left(6x+12\right)

$(x_{2}+2x)+(6x+12)$

Factor out x in the first and 6 in the second group.

Factor out $x$ in the first and $6$ in the second group.

x\left(x+2\right)+6\left(x+2\right)

$x(x+2)+6(x+2)$

Factor out common term x+2 by using distributive property.

Factor out common term $x+2$ by using distributive property.

\left(x+2\right)\left(x+6\right)

$(x+2)(x+6)$

Rewrite the complete factored expression.

Rewrite the complete factored expression.

2\left(x+2\right)\left(x+6\right)

$2(x+2)(x+6)$

Evaluate

2\left(x+2\right)\left(x+6\right)

$2(x+2)(x+6)$

Graph

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2\left(x^{2}+8x+12\right)

Factor out 2.

a+b=8 ab=1\times 12=12

Consider x^{2}+8x+12. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

1,12 2,6 3,4

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

1+12=13 2+6=8 3+4=7

Calculate the sum for each pair.

a=2 b=6

The solution is the pair that gives sum 8.

\left(x^{2}+2x\right)+\left(6x+12\right)

Rewrite x^{2}+8x+12 as \left(x^{2}+2x\right)+\left(6x+12\right).

x\left(x+2\right)+6\left(x+2\right)

Factor out x in the first and 6 in the second group.

\left(x+2\right)\left(x+6\right)

Factor out common term x+2 by using distributive property.

2\left(x+2\right)\left(x+6\right)

Rewrite the complete factored expression.

2x^{2}+16x+24=0

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

x=\frac{-16±\sqrt{16^{2}-4\times 2\times 24}}{2\times 2}

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

x=\frac{-16±\sqrt{256-4\times 2\times 24}}{2\times 2}

Square 16.

x=\frac{-16±\sqrt{256-8\times 24}}{2\times 2}

Multiply -4 times 2.

x=\frac{-16±\sqrt{256-192}}{2\times 2}

Multiply -8 times 24.

x=\frac{-16±\sqrt{64}}{2\times 2}

Add 256 to -192.

x=\frac{-16±8}{2\times 2}

Take the square root of 64.

x=\frac{-16±8}{4}

Multiply 2 times 2.

x=\frac{-8}{4}

Now solve the equation x=\frac{-16±8}{4} when ± is plus. Add -16 to 8.

x=-2

Divide -8 by 4.

x=\frac{-24}{4}

Now solve the equation x=\frac{-16±8}{4} when ± is minus. Subtract 8 from -16.

x=-6

Divide -24 by 4.

2x^{2}+16x+24=2\left(x-\left(-2\right)\right)\left(x-\left(-6\right)\right)

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

2x^{2}+16x+24=2\left(x+2\right)\left(x+6\right)

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

x ^ 2 +8x +12 = 0

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 2

r + s = -8 rs = 12

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

r = -4 - u s = -4 + u

Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -4. 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 style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(-4 - u) (-4 + u) = 12

To solve for unknown quantity u, substitute these in the product equation rs = 12

16 - u^2 = 12

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

-u^2 = 12-16 = -4

Simplify the expression by subtracting 16 on both sides

u^2 = 4 u = \pm\sqrt{4} = \pm 2

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r =-4 - 2 = -6 s = -4 + 2 = -2

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

Examples

Quadratic equation

{ x } ^ { 2 } - 4 x - 5 = 0

$x_{2}−4x−5=0$

Trigonometry

4 \sin \theta \cos \theta = 2 \sin \theta

$4sinθcosθ=2sinθ$

Linear equation

y = 3x + 4

$y=3x+4$

Arithmetic

699 * 533

$699∗533$

Matrix

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \right]

$[25 34 ][2−1 01 35 ]$

Simultaneous equation

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

${8x+2y=467x+3y=47 $

Differentiation

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

$dxd (x−5)(3x_{2}−2) $

Integration

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

$∫_{0}xe_{−x_{2}}dx$

Limits

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

$x→−3lim x_{2}+2x−3x_{2}−9 $

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