Solve -3x^2+4x+15

$$-3 { x }^{ 2 } +4x+15$$

Factor

$$\left(3-x\right)\left(3x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=4 ab=-3\times 15=-45$$

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 -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=9 b=-5$$

Rewrite -3x^{2}+4x+15 as \left(-3x^{2}+9x\right)+\left(-5x+15\right).

$$\left(-3x^{2}+9x\right)+\left(-5x+15\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(-x+3\right)+5\left(-x+3\right)$$

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

$$\left(-x+3\right)\left(3x+5\right)$$

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.

$$-3x^{2}+4x+15=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-3\right)\times 15}}{2\left(-3\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-3\right)\times 15}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-4±\sqrt{16+12\times 15}}{2\left(-3\right)}$$

Multiply 12 times 15.

$$x=\frac{-4±\sqrt{16+180}}{2\left(-3\right)}$$

Add 16 to 180.

$$x=\frac{-4±\sqrt{196}}{2\left(-3\right)}$$

Take the square root of 196.

$$x=\frac{-4±14}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-4±14}{-6}$$

Now solve the equation x=\frac{-4±14}{-6} when ± is plus. Add -4 to 14.

$$x=\frac{10}{-6}$$

Reduce the fraction \frac{10}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{3}$$

Now solve the equation x=\frac{-4±14}{-6} when ± is minus. Subtract 14 from -4.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{3} for x_{1} and 3 for x_{2}.

$$-3x^{2}+4x+15=-3\left(x-\left(-\frac{5}{3}\right)\right)\left(x-3\right)$$

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

$$-3x^{2}+4x+15=-3\left(x+\frac{5}{3}\right)\left(x-3\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-3x^{2}+4x+15=-3\times \frac{-3x-5}{-3}\left(x-3\right)$$

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

$$-3x^{2}+4x+15=\left(-3x-5\right)\left(x-3\right)$$

Evaluate

$$\left(3-x\right)\left(3x+5\right)$$