Solve 4x^2+10x+6

$$4 x ^ { 2 } + 10 x + 6$$

Factor

$$2\left(x+1\right)\left(2x+3\right)$$

Solution Steps

Factor out 2.

$$2\left(2x^{2}+5x+3\right)$$

Consider 2x^{2}+5x+3. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=5 ab=2\times 3=6$$

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=2 b=3$$

Rewrite 2x^{2}+5x+3 as \left(2x^{2}+2x\right)+\left(3x+3\right).

$$\left(2x^{2}+2x\right)+\left(3x+3\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(x+1\right)+3\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(2x+3\right)$$

Rewrite the complete factored expression.

$$2\left(x+1\right)\left(2x+3\right)$$

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.

$$4x^{2}+10x+6=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 4\times 6}}{2\times 4}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-10±\sqrt{100-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$x=\frac{-10±\sqrt{100-96}}{2\times 4}$$

Add 100 to -96.

$$x=\frac{-10±\sqrt{4}}{2\times 4}$$

Take the square root of 4.

$$x=\frac{-10±2}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-10±2}{8}$$

Now solve the equation x=\frac{-10±2}{8} when ± is plus. Add -10 to 2.

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

Divide -8 by 8.

$$x=-1$$

Now solve the equation x=\frac{-10±2}{8} when ± is minus. Subtract 2 from -10.

$$x=-\frac{12}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and -\frac{3}{2} for x_{2}.

$$4x^{2}+10x+6=4\left(x-\left(-1\right)\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$4x^{2}+10x+6=4\left(x+1\right)\left(x+\frac{3}{2}\right)$$

Add \frac{3}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+10x+6=4\left(x+1\right)\times \frac{2x+3}{2}$$

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

$$4x^{2}+10x+6=2\left(x+1\right)\left(2x+3\right)$$

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

$$x ^ 2 +\frac{5}{2}x +\frac{3}{2} = 0$$

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

$$r + s = -\frac{5}{2} rs = \frac{3}{2}$$

Two numbers r and s sum up to -\frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{2} = -\frac{5}{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>

$$r = -\frac{5}{4} - u s = -\frac{5}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{2}

$$(-\frac{5}{4} - u) (-\frac{5}{4} + u) = \frac{3}{2}$$

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

$$\frac{25}{16} - u^2 = \frac{3}{2}$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = \frac{3}{2}-\frac{25}{16} = -\frac{1}{16}$$

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

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

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

$$r =-\frac{5}{4} - \frac{1}{4} = -1.500 s = -\frac{5}{4} + \frac{1}{4} = -1$$

Evaluate

$$2\left(x+1\right)\left(2x+3\right)$$