Solve 14x^2+x-3

$$14 { x }^{ 2 } +x-3$$

Factor

$$\left(7x-3\right)\left(2x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 14x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=1 ab=14\left(-3\right)=-42$$

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 -42.

$$-1,42 -2,21 -3,14 -6,7$$

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 1.

$$a=-6 b=7$$

Rewrite 14x^{2}+x-3 as \left(14x^{2}-6x\right)+\left(7x-3\right).

$$\left(14x^{2}-6x\right)+\left(7x-3\right)$$

Factor out 2x in 14x^{2}-6x.

$$2x\left(7x-3\right)+7x-3$$

Factor out common term 7x-3 by using distributive property.

$$\left(7x-3\right)\left(2x+1\right)$$

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.

$$14x^{2}+x-3=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 14\left(-3\right)}}{2\times 14}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 14\left(-3\right)}}{2\times 14}$$

Multiply -4 times 14.

$$x=\frac{-1±\sqrt{1-56\left(-3\right)}}{2\times 14}$$

Multiply -56 times -3.

$$x=\frac{-1±\sqrt{1+168}}{2\times 14}$$

Add 1 to 168.

$$x=\frac{-1±\sqrt{169}}{2\times 14}$$

Take the square root of 169.

$$x=\frac{-1±13}{2\times 14}$$

Multiply 2 times 14.

$$x=\frac{-1±13}{28}$$

Now solve the equation x=\frac{-1±13}{28} when ± is plus. Add -1 to 13.

$$x=\frac{12}{28}$$

Reduce the fraction \frac{12}{28} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{7}$$

Now solve the equation x=\frac{-1±13}{28} when ± is minus. Subtract 13 from -1.

$$x=-\frac{14}{28}$$

Reduce the fraction \frac{-14}{28} to lowest terms by extracting and canceling out 14.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{7} for x_{1} and -\frac{1}{2} for x_{2}.

$$14x^{2}+x-3=14\left(x-\frac{3}{7}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$14x^{2}+x-3=14\left(x-\frac{3}{7}\right)\left(x+\frac{1}{2}\right)$$

Subtract \frac{3}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$14x^{2}+x-3=14\times \frac{7x-3}{7}\left(x+\frac{1}{2}\right)$$

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

$$14x^{2}+x-3=14\times \frac{7x-3}{7}\times \frac{2x+1}{2}$$

Multiply \frac{7x-3}{7} times \frac{2x+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$14x^{2}+x-3=14\times \frac{\left(7x-3\right)\left(2x+1\right)}{7\times 2}$$

Multiply 7 times 2.

$$14x^{2}+x-3=14\times \frac{\left(7x-3\right)\left(2x+1\right)}{14}$$

Cancel out 14, the greatest common factor in 14 and 14.

$$14x^{2}+x-3=\left(7x-3\right)\left(2x+1\right)$$

Evaluate

$$14x^{2}+x-3$$