Solve 27+30x-25x^2

$$27 + 30 x - 25 x ^ { 2 }$$

Factor

$$-\left(5x-9\right)\left(5x+3\right)$$

Solution Steps

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

$$-25x^{2}+30x+27$$

Factor the expression by grouping. First, the expression needs to be rewritten as -25x^{2}+ax+bx+27. To find a and b, set up a system to be solved.

$$a+b=30 ab=-25\times 27=-675$$

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 -675.

$$-1,675 -3,225 -5,135 -9,75 -15,45 -25,27$$

Calculate the sum for each pair.

$$-1+675=674 -3+225=222 -5+135=130 -9+75=66 -15+45=30 -25+27=2$$

The solution is the pair that gives sum 30.

$$a=45 b=-15$$

Rewrite -25x^{2}+30x+27 as \left(-25x^{2}+45x\right)+\left(-15x+27\right).

$$\left(-25x^{2}+45x\right)+\left(-15x+27\right)$$

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

$$-5x\left(5x-9\right)-3\left(5x-9\right)$$

Factor out common term 5x-9 by using distributive property.

$$\left(5x-9\right)\left(-5x-3\right)$$

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.

$$-25x^{2}+30x+27=0$$

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.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-25\right)\times 27}}{2\left(-25\right)}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\left(-25\right)\times 27}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-30±\sqrt{900+100\times 27}}{2\left(-25\right)}$$

Multiply 100 times 27.

$$x=\frac{-30±\sqrt{900+2700}}{2\left(-25\right)}$$

Add 900 to 2700.

$$x=\frac{-30±\sqrt{3600}}{2\left(-25\right)}$$

Take the square root of 3600.

$$x=\frac{-30±60}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{-30±60}{-50}$$

Now solve the equation x=\frac{-30±60}{-50} when ± is plus. Add -30 to 60.

$$x=\frac{30}{-50}$$

Reduce the fraction \frac{30}{-50} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{-30±60}{-50} when ± is minus. Subtract 60 from -30.

$$x=-\frac{90}{-50}$$

Reduce the fraction \frac{-90}{-50} to lowest terms by extracting and canceling out 10.

$$x=\frac{9}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{5} for x_{1} and \frac{9}{5} for x_{2}.

$$-25x^{2}+30x+27=-25\left(x-\left(-\frac{3}{5}\right)\right)\left(x-\frac{9}{5}\right)$$

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

$$-25x^{2}+30x+27=-25\left(x+\frac{3}{5}\right)\left(x-\frac{9}{5}\right)$$

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

$$-25x^{2}+30x+27=-25\times \frac{-5x-3}{-5}\left(x-\frac{9}{5}\right)$$

Subtract \frac{9}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-25x^{2}+30x+27=-25\times \frac{-5x-3}{-5}\times \frac{-5x+9}{-5}$$

Multiply \frac{-5x-3}{-5} times \frac{-5x+9}{-5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-25x^{2}+30x+27=-25\times \frac{\left(-5x-3\right)\left(-5x+9\right)}{-5\left(-5\right)}$$

Multiply -5 times -5.

$$-25x^{2}+30x+27=-25\times \frac{\left(-5x-3\right)\left(-5x+9\right)}{25}$$

Cancel out 25, the greatest common factor in -25 and 25.

$$-25x^{2}+30x+27=-\left(-5x-3\right)\left(-5x+9\right)$$

Evaluate

$$27+30x-25x^{2}$$