Solve 3x^2+x-4

$$3 { x }^{ 2 } +x-4$$

Factor

$$\left(x-1\right)\left(3x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=1 ab=3\left(-4\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite 3x^{2}+x-4 as \left(3x^{2}-3x\right)+\left(4x-4\right).

$$\left(3x^{2}-3x\right)+\left(4x-4\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(x-1\right)+4\left(x-1\right)$$

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

$$\left(x-1\right)\left(3x+4\right)$$

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.

$$3x^{2}+x-4=0$$

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.

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

Square 1.

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

Multiply -4 times 3.

$$x=\frac{-1±\sqrt{1-12\left(-4\right)}}{2\times 3}$$

Multiply -12 times -4.

$$x=\frac{-1±\sqrt{1+48}}{2\times 3}$$

Add 1 to 48.

$$x=\frac{-1±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$x=\frac{-1±7}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-1±7}{6}$$

Now solve the equation x=\frac{-1±7}{6} when ± is plus. Add -1 to 7.

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

Divide 6 by 6.

$$x=1$$

Now solve the equation x=\frac{-1±7}{6} when ± is minus. Subtract 7 from -1.

$$x=-\frac{8}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{3}$$

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{4}{3} for x_{2}.

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

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

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

Add \frac{4}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Cancel out 3, the greatest common factor in 3 and 3.

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

Evaluate

$$\left(x-1\right)\left(3x+4\right)$$