Solve 2x^2+19x+45

$$2 { x }^{ 2 } +19x+45$$

Factor

$$\left(x+5\right)\left(2x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=19 ab=2\times 45=90$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 90.

$$1,90 2,45 3,30 5,18 6,15 9,10$$

Calculate the sum for each pair.

$$1+90=91 2+45=47 3+30=33 5+18=23 6+15=21 9+10=19$$

The solution is the pair that gives sum 19.

$$a=9 b=10$$

Rewrite 2x^{2}+19x+45 as \left(2x^{2}+9x\right)+\left(10x+45\right).

$$\left(2x^{2}+9x\right)+\left(10x+45\right)$$

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

$$x\left(2x+9\right)+5\left(2x+9\right)$$

Factor out common term 2x+9 by using distributive property.

$$\left(2x+9\right)\left(x+5\right)$$

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.

$$2x^{2}+19x+45=0$$

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.

$$x=\frac{-19±\sqrt{19^{2}-4\times 2\times 45}}{2\times 2}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\times 2\times 45}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-19±\sqrt{361-8\times 45}}{2\times 2}$$

Multiply -8 times 45.

$$x=\frac{-19±\sqrt{361-360}}{2\times 2}$$

Add 361 to -360.

$$x=\frac{-19±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$x=\frac{-19±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-19±1}{4}$$

Now solve the equation x=\frac{-19±1}{4} when ± is plus. Add -19 to 1.

$$x=-\frac{18}{4}$$

Reduce the fraction \frac{-18}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{9}{2}$$

Now solve the equation x=\frac{-19±1}{4} when ± is minus. Subtract 1 from -19.

$$x=-\frac{20}{4}$$

Divide -20 by 4.

$$x=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{9}{2} for x_{1} and -5 for x_{2}.

$$2x^{2}+19x+45=2\left(x-\left(-\frac{9}{2}\right)\right)\left(x-\left(-5\right)\right)$$

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

$$2x^{2}+19x+45=2\left(x+\frac{9}{2}\right)\left(x+5\right)$$

Add \frac{9}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$2x^{2}+19x+45=2\times \frac{2x+9}{2}\left(x+5\right)$$

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

$$2x^{2}+19x+45=\left(2x+9\right)\left(x+5\right)$$

Evaluate

$$\left(x+5\right)\left(2x+9\right)$$