Solve F(x)=22x^2+49x-15

$$F ( x ) = 22 x ^ { 2 } + 49 x - 15$$

Factor

$$\left(11x-3\right)\left(2x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 22x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=49 ab=22\left(-15\right)=-330$$

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 -330.

$$-1,330 -2,165 -3,110 -5,66 -6,55 -10,33 -11,30 -15,22$$

Calculate the sum for each pair.

$$-1+330=329 -2+165=163 -3+110=107 -5+66=61 -6+55=49 -10+33=23 -11+30=19 -15+22=7$$

The solution is the pair that gives sum 49.

$$a=-6 b=55$$

Rewrite 22x^{2}+49x-15 as \left(22x^{2}-6x\right)+\left(55x-15\right).

$$\left(22x^{2}-6x\right)+\left(55x-15\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(11x-3\right)+5\left(11x-3\right)$$

Factor out common term 11x-3 by using distributive property.

$$\left(11x-3\right)\left(2x+5\right)$$

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.

$$22x^{2}+49x-15=0$$

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.

$$x=\frac{-49±\sqrt{49^{2}-4\times 22\left(-15\right)}}{2\times 22}$$

Square 49.

$$x=\frac{-49±\sqrt{2401-4\times 22\left(-15\right)}}{2\times 22}$$

Multiply -4 times 22.

$$x=\frac{-49±\sqrt{2401-88\left(-15\right)}}{2\times 22}$$

Multiply -88 times -15.

$$x=\frac{-49±\sqrt{2401+1320}}{2\times 22}$$

Add 2401 to 1320.

$$x=\frac{-49±\sqrt{3721}}{2\times 22}$$

Take the square root of 3721.

$$x=\frac{-49±61}{2\times 22}$$

Multiply 2 times 22.

$$x=\frac{-49±61}{44}$$

Now solve the equation x=\frac{-49±61}{44} when ± is plus. Add -49 to 61.

$$x=\frac{12}{44}$$

Reduce the fraction \frac{12}{44} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{11}$$

Now solve the equation x=\frac{-49±61}{44} when ± is minus. Subtract 61 from -49.

$$x=-\frac{110}{44}$$

Reduce the fraction \frac{-110}{44} to lowest terms by extracting and canceling out 22.

$$x=-\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{11} for x_{1} and -\frac{5}{2} for x_{2}.

$$22x^{2}+49x-15=22\left(x-\frac{3}{11}\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$22x^{2}+49x-15=22\left(x-\frac{3}{11}\right)\left(x+\frac{5}{2}\right)$$

Subtract \frac{3}{11} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$22x^{2}+49x-15=22\times \frac{11x-3}{11}\left(x+\frac{5}{2}\right)$$

Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$22x^{2}+49x-15=22\times \frac{11x-3}{11}\times \frac{2x+5}{2}$$

Multiply \frac{11x-3}{11} times \frac{2x+5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$22x^{2}+49x-15=22\times \frac{\left(11x-3\right)\left(2x+5\right)}{11\times 2}$$

Multiply 11 times 2.

$$22x^{2}+49x-15=22\times \frac{\left(11x-3\right)\left(2x+5\right)}{22}$$

Cancel out 22, the greatest common factor in 22 and 22.

$$22x^{2}+49x-15=\left(11x-3\right)\left(2x+5\right)$$

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 22

$$x ^ 2 +\frac{49}{22}x -\frac{15}{22} = 0$$

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

$$r + s = -\frac{49}{22} rs = -\frac{15}{22}$$

Two numbers r and s sum up to -\frac{49}{22} exactly when the average of the two numbers is \frac{1}{2}*-\frac{49}{22} = -\frac{49}{44}. 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>

$$r = -\frac{49}{44} - u s = -\frac{49}{44} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{15}{22}

$$(-\frac{49}{44} - u) (-\frac{49}{44} + u) = -\frac{15}{22}$$

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

$$\frac{2401}{1936} - u^2 = -\frac{15}{22}$$

Simplify the expression by subtracting \frac{2401}{1936} on both sides

$$-u^2 = -\frac{15}{22}-\frac{2401}{1936} = -\frac{3721}{1936}$$

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

$$u^2 = \frac{3721}{1936} u = \pm\sqrt{\frac{3721}{1936}} = \pm \frac{61}{44}$$

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

$$r =-\frac{49}{44} - \frac{61}{44} = -2.500 s = -\frac{49}{44} + \frac{61}{44} = 0.273$$

Evaluate

$$\left(11x-3\right)\left(2x+5\right)$$