Solve 2x+5-3x^2

$$2 x + 5 - 3 x ^ { 2 }$$

Factor

$$-\left(3x-5\right)\left(x+1\right)$$

Solution Steps

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

$$-3x^{2}+2x+5$$

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3\times 5=-15$$

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 -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=5 b=-3$$

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

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

Factor out -x in the first and -1 in the second group.

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

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(-x-1\right)$$

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.

$$-3x^{2}+2x+5=0$$

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.

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

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-3\right)\times 5}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-2±\sqrt{4+12\times 5}}{2\left(-3\right)}$$

Multiply 12 times 5.

$$x=\frac{-2±\sqrt{4+60}}{2\left(-3\right)}$$

Add 4 to 60.

$$x=\frac{-2±\sqrt{64}}{2\left(-3\right)}$$

Take the square root of 64.

$$x=\frac{-2±8}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-2±8}{-6}$$

Now solve the equation x=\frac{-2±8}{-6} when ± is plus. Add -2 to 8.

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

Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{-2±8}{-6} when ± is minus. Subtract 8 from -2.

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

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

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

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{5}{3} for x_{2}.

$$-3x^{2}+2x+5=-3\left(x-\left(-1\right)\right)\left(x-\frac{5}{3}\right)$$

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

$$-3x^{2}+2x+5=-3\left(x+1\right)\left(x-\frac{5}{3}\right)$$

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

$$-3x^{2}+2x+5=-3\left(x+1\right)\times \frac{-3x+5}{-3}$$

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

$$-3x^{2}+2x+5=\left(x+1\right)\left(-3x+5\right)$$

Evaluate

$$-\left(3x-5\right)\left(x+1\right)$$