Solve f(x)=10x^2+23x-42

$$f ( x ) = 10 x ^ { 2 } + 23 x - 42$$

Factor

$$\left(5x-6\right)\left(2x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-42. To find a and b, set up a system to be solved.

$$a+b=23 ab=10\left(-42\right)=-420$$

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 -420.

$$-1,420 -2,210 -3,140 -4,105 -5,84 -6,70 -7,60 -10,42 -12,35 -14,30 -15,28 -20,21$$

Calculate the sum for each pair.

$$-1+420=419 -2+210=208 -3+140=137 -4+105=101 -5+84=79 -6+70=64 -7+60=53 -10+42=32 -12+35=23 -14+30=16 -15+28=13 -20+21=1$$

The solution is the pair that gives sum 23.

$$a=-12 b=35$$

Rewrite 10x^{2}+23x-42 as \left(10x^{2}-12x\right)+\left(35x-42\right).

$$\left(10x^{2}-12x\right)+\left(35x-42\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(5x-6\right)+7\left(5x-6\right)$$

Factor out common term 5x-6 by using distributive property.

$$\left(5x-6\right)\left(2x+7\right)$$

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.

$$10x^{2}+23x-42=0$$

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.

$$x=\frac{-23±\sqrt{23^{2}-4\times 10\left(-42\right)}}{2\times 10}$$

Square 23.

$$x=\frac{-23±\sqrt{529-4\times 10\left(-42\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-23±\sqrt{529-40\left(-42\right)}}{2\times 10}$$

Multiply -40 times -42.

$$x=\frac{-23±\sqrt{529+1680}}{2\times 10}$$

Add 529 to 1680.

$$x=\frac{-23±\sqrt{2209}}{2\times 10}$$

Take the square root of 2209.

$$x=\frac{-23±47}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-23±47}{20}$$

Now solve the equation x=\frac{-23±47}{20} when ± is plus. Add -23 to 47.

$$x=\frac{24}{20}$$

Reduce the fraction \frac{24}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{6}{5}$$

Now solve the equation x=\frac{-23±47}{20} when ± is minus. Subtract 47 from -23.

$$x=-\frac{70}{20}$$

Reduce the fraction \frac{-70}{20} to lowest terms by extracting and canceling out 10.

$$x=-\frac{7}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{6}{5} for x_{1} and -\frac{7}{2} for x_{2}.

$$10x^{2}+23x-42=10\left(x-\frac{6}{5}\right)\left(x-\left(-\frac{7}{2}\right)\right)$$

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

$$10x^{2}+23x-42=10\left(x-\frac{6}{5}\right)\left(x+\frac{7}{2}\right)$$

Subtract \frac{6}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+23x-42=10\times \frac{5x-6}{5}\left(x+\frac{7}{2}\right)$$

Add \frac{7}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+23x-42=10\times \frac{5x-6}{5}\times \frac{2x+7}{2}$$

Multiply \frac{5x-6}{5} times \frac{2x+7}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+23x-42=10\times \frac{\left(5x-6\right)\left(2x+7\right)}{5\times 2}$$

Multiply 5 times 2.

$$10x^{2}+23x-42=10\times \frac{\left(5x-6\right)\left(2x+7\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10x^{2}+23x-42=\left(5x-6\right)\left(2x+7\right)$$

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 10

$$x ^ 2 +\frac{23}{10}x -\frac{21}{5} = 0$$

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

$$r + s = -\frac{23}{10} rs = -\frac{21}{5}$$

Two numbers r and s sum up to -\frac{23}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{23}{10} = -\frac{23}{20}. 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>

$$r = -\frac{23}{20} - u s = -\frac{23}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{21}{5}

$$(-\frac{23}{20} - u) (-\frac{23}{20} + u) = -\frac{21}{5}$$

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

$$\frac{529}{400} - u^2 = -\frac{21}{5}$$

Simplify the expression by subtracting \frac{529}{400} on both sides

$$-u^2 = -\frac{21}{5}-\frac{529}{400} = -\frac{2209}{400}$$

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

$$u^2 = \frac{2209}{400} u = \pm\sqrt{\frac{2209}{400}} = \pm \frac{47}{20}$$

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

$$r =-\frac{23}{20} - \frac{47}{20} = -3.500 s = -\frac{23}{20} + \frac{47}{20} = 1.200$$

Evaluate

$$\left(5x-6\right)\left(2x+7\right)$$