Solve 2x^2-3+x

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

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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.

$$a+b=1 ab=2\left(-3\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

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

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite 2x^{2}+x-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.

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

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.

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

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.

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

Square 1.

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

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-3\right)}}{2\times 2}$$

Multiply -8 times -3.

$$x=\frac{-1±\sqrt{1+24}}{2\times 2}$$

Add 1 to 24.

$$x=\frac{-1±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-1±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±5}{4}$$

Now solve the equation x=\frac{-1±5}{4} when ± is plus. Add -1 to 5.

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

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{-1±5}{4} when ± is minus. Subtract 5 from -1.

$$x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

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

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}.

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

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

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

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

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

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

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

Evaluate

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