Solve 4x^2+19x-30

$$4 { x }^{ 2 } +19x-30$$

Factor

$$\left(4x-5\right)\left(x+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=19 ab=4\left(-30\right)=-120$$

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 -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 19.

$$a=-5 b=24$$

Rewrite 4x^{2}+19x-30 as \left(4x^{2}-5x\right)+\left(24x-30\right).

$$\left(4x^{2}-5x\right)+\left(24x-30\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(4x-5\right)+6\left(4x-5\right)$$

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(x+6\right)$$

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.

$$4x^{2}+19x-30=0$$

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.

$$x=\frac{-19±\sqrt{19^{2}-4\times 4\left(-30\right)}}{2\times 4}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\times 4\left(-30\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-19±\sqrt{361-16\left(-30\right)}}{2\times 4}$$

Multiply -16 times -30.

$$x=\frac{-19±\sqrt{361+480}}{2\times 4}$$

Add 361 to 480.

$$x=\frac{-19±\sqrt{841}}{2\times 4}$$

Take the square root of 841.

$$x=\frac{-19±29}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-19±29}{8}$$

Now solve the equation x=\frac{-19±29}{8} when ± is plus. Add -19 to 29.

$$x=\frac{10}{8}$$

Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{4}$$

Now solve the equation x=\frac{-19±29}{8} when ± is minus. Subtract 29 from -19.

$$x=-\frac{48}{8}$$

Divide -48 by 8.

$$x=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{4} for x_{1} and -6 for x_{2}.

$$4x^{2}+19x-30=4\left(x-\frac{5}{4}\right)\left(x-\left(-6\right)\right)$$

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

$$4x^{2}+19x-30=4\left(x-\frac{5}{4}\right)\left(x+6\right)$$

Subtract \frac{5}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+19x-30=4\times \frac{4x-5}{4}\left(x+6\right)$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4x^{2}+19x-30=\left(4x-5\right)\left(x+6\right)$$

Evaluate

$$\left(4x-5\right)\left(x+6\right)$$