3x To The Second Power

$iii \exponential{x}{2} - 10 ten + eight $

\left(x-2\right)\left(3x-4\right)

\left(10-ii\right)\left(3x-4\right)

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a+b=-10 ab=3\times 8=24

Gene the expression past grouping. First, the expression needs to be rewritten as 3x^{ii}+ax+bx+8. To notice a and b, set up a arrangement to exist solved.

-1,-24 -2,-12 -3,-viii -four,-half-dozen

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. Listing all such integer pairs that requite product 24.

-one-24=-25 -two-12=-14 -iii-8=-11 -4-half dozen=-10

Calculate the sum for each pair.

a=-6 b=-four

The solution is the pair that gives sum -10.

\left(3x^{2}-6x\right)+\left(-4x+viii\correct)

Rewrite 3x^{2}-10x+viii as \left(3x^{two}-6x\right)+\left(-4x+8\right).

3x\left(x-ii\right)-4\left(x-two\right)

Factor out 3x in the starting time and -4 in the second group.

\left(x-ii\right)\left(3x-iv\right)

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

3x^{2}-10x+viii=0

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

ten=\frac{-\left(-10\correct)±\sqrt{\left(-10\right)^{2}-4\times 3\times 8}}{2\times iii}

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

x=\frac{-\left(-ten\correct)±\sqrt{100-4\times 3\times 8}}{2\times 3}

Square -ten.

x=\frac{-\left(-10\right)±\sqrt{100-12\times 8}}{two\times 3}

Multiply -four times 3.

x=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times three}

Multiply -12 times 8.

x=\frac{-\left(-ten\right)±\sqrt{4}}{2\times three}

Add 100 to -96.

x=\frac{-\left(-10\correct)±2}{2\times 3}

Take the square root of 4.

10=\frac{x±2}{2\times 3}

The opposite of -10 is ten.

x=\frac{10±2}{half-dozen}

Multiply two times three.

x=\frac{12}{half-dozen}

At present solve the equation x=\frac{10±2}{vi} when ± is plus. Add together ten to 2.

x=\frac{8}{six}

At present solve the equation x=\frac{10±two}{6} when ± is minus. Subtract 2 from 10.

ten=\frac{four}{3}

Reduce the fraction \frac{8}{6} to everyman terms by extracting and canceling out 2.

3x^{2}-10x+8=3\left(ten-ii\right)\left(x-\frac{4}{3}\right)

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

3x^{2}-10x+8=3\left(ten-2\correct)\times \left(\frac{3x-4}{3}\right)

Subtract \frac{4}{three} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

3x^{two}-10x+eight=\left(x-2\right)\left(3x-four\right)

Abolish out iii, the greatest mutual cistron in iii and 3.

x ^ 2 -\frac{10}{3}10 +\frac{viii}{3} = 0

Quadratic equations such every bit this i tin be solved past a new straight factoring method that does non crave guess work. To use the direct factoring method, the equation must be in the form 10^2+Bx+C=0.This is achieved by dividing both sides of the equation by three

r + s = \frac{10}{3} rs = \frac{8}{3}

Let r and s be the factors for the quadratic equation such that 10^2+Bx+C=(x−r)(x−s) where sum of factors (r+due south)=−B and the product of factors rs = C

r = \frac{5}{3} - u s = \frac{5}{3} + u

Two numbers r and s sum up to \frac{10}{three} exactly when the average of the 2 numbers is \frac{1}{2}*\frac{x}{3} = \frac{five}{3}. You can also come across that the midpoint of r and s corresponds to the centrality of symmetry of the parabola represented past the quadratic equation y=x^2+Bx+C. The values of r and due south are equidistant from the heart by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.internet/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(\frac{v}{iii} - u) (\frac{5}{three} + u) = \frac{8}{3}

To solve for unknown quantity u, substitute these in the production equation rs = \frac{8}{3}

\frac{25}{9} - u^2 = \frac{8}{3}

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

-u^2 = \frac{viii}{3}-\frac{25}{9} = -\frac{ane}{9}

Simplify the expression by subtracting \frac{25}{nine} on both sides

u^2 = \frac{1}{9} u = \pm\sqrt{\frac{one}{nine}} = \pm \frac{1}{3}

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

r =\frac{5}{3} - \frac{ane}{iii} = 1.333 southward = \frac{5}{3} + \frac{1}{3} = 2.000

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

3x To The Second Power,

Source: https://mathsolver.microsoft.com/en/topic/algebra/factor/solve/3x%5E2-10x+8

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