Solve 4x^2+18x-10=

$$4 x ^ { 2 } + 18 x - 10 =$$

Factor

$$2\left(2x-1\right)\left(x+5\right)$$

Solution Steps

Factor out 2.

$$2\left(2x^{2}+9x-5\right)$$

Consider 2x^{2}+9x-5. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=9 ab=2\left(-5\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 9.

$$a=-1 b=10$$

Rewrite 2x^{2}+9x-5 as \left(2x^{2}-x\right)+\left(10x-5\right).

$$\left(2x^{2}-x\right)+\left(10x-5\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(2x-1\right)+5\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(x+5\right)$$

Rewrite the complete factored expression.

$$2\left(2x-1\right)\left(x+5\right)$$

Steps Using the Quadratic Formula

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.

$$4x^{2}+18x-10=0$$

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.

$$x=\frac{-18±\sqrt{18^{2}-4\times 4\left(-10\right)}}{2\times 4}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 4\left(-10\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-18±\sqrt{324-16\left(-10\right)}}{2\times 4}$$

Multiply -16 times -10.

$$x=\frac{-18±\sqrt{324+160}}{2\times 4}$$

Add 324 to 160.

$$x=\frac{-18±\sqrt{484}}{2\times 4}$$

Take the square root of 484.

$$x=\frac{-18±22}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-18±22}{8}$$

Now solve the equation x=\frac{-18±22}{8} when ± is plus. Add -18 to 22.

$$x=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-18±22}{8} when ± is minus. Subtract 22 from -18.

$$x=-\frac{40}{8}$$

Divide -40 by 8.

$$x=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2} for x_{1} and -5 for x_{2}.

$$4x^{2}+18x-10=4\left(x-\frac{1}{2}\right)\left(x-\left(-5\right)\right)$$

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

$$4x^{2}+18x-10=4\left(x-\frac{1}{2}\right)\left(x+5\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+18x-10=4\times \frac{2x-1}{2}\left(x+5\right)$$

Cancel out 2, the greatest common factor in 4 and 2.

$$4x^{2}+18x-10=2\left(2x-1\right)\left(x+5\right)$$

Steps Using Direct Factoring Method

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 4

$$x ^ 2 +\frac{9}{2}x -\frac{5}{2} = 0$$

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

$$r + s = -\frac{9}{2} rs = -\frac{5}{2}$$

Two numbers r and s sum up to -\frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{9}{2} = -\frac{9}{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>

$$r = -\frac{9}{4} - u s = -\frac{9}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}

$$(-\frac{9}{4} - u) (-\frac{9}{4} + u) = -\frac{5}{2}$$

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

$$\frac{81}{16} - u^2 = -\frac{5}{2}$$

Simplify the expression by subtracting \frac{81}{16} on both sides

$$-u^2 = -\frac{5}{2}-\frac{81}{16} = -\frac{121}{16}$$

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

$$u^2 = \frac{121}{16} u = \pm\sqrt{\frac{121}{16}} = \pm \frac{11}{4}$$

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

$$r =-\frac{9}{4} - \frac{11}{4} = -5 s = -\frac{9}{4} + \frac{11}{4} = 0.500$$

Evaluate

$$2\left(2x-1\right)\left(x+5\right)$$