Solve -4x^2-5x+9

$$-4 { x }^{ 2 } -5x+9$$

Factor

$$\left(1-x\right)\left(4x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -4x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-4\times 9=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -5.

$$a=4 b=-9$$

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

$$\left(-4x^{2}+4x\right)+\left(-9x+9\right)$$

Factor out 4x in the first and 9 in the second group.

$$4x\left(-x+1\right)+9\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(4x+9\right)$$

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.

$$-4x^{2}-5x+9=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-4\right)\times 9}}{2\left(-4\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-4\right)\times 9}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-5\right)±\sqrt{25+16\times 9}}{2\left(-4\right)}$$

Multiply 16 times 9.

$$x=\frac{-\left(-5\right)±\sqrt{25+144}}{2\left(-4\right)}$$

Add 25 to 144.

$$x=\frac{-\left(-5\right)±\sqrt{169}}{2\left(-4\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-5\right)±13}{2\left(-4\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±13}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{5±13}{-8}$$

Now solve the equation x=\frac{5±13}{-8} when ± is plus. Add 5 to 13.

$$x=\frac{18}{-8}$$

Reduce the fraction \frac{18}{-8} to lowest terms by extracting and canceling out 2.

$$x=-\frac{9}{4}$$

Now solve the equation x=\frac{5±13}{-8} when ± is minus. Subtract 13 from 5.

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

Divide -8 by -8.

$$x=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{9}{4} for x_{1} and 1 for x_{2}.

$$-4x^{2}-5x+9=-4\left(x-\left(-\frac{9}{4}\right)\right)\left(x-1\right)$$

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

$$-4x^{2}-5x+9=-4\left(x+\frac{9}{4}\right)\left(x-1\right)$$

Add \frac{9}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}-5x+9=-4\times \frac{-4x-9}{-4}\left(x-1\right)$$

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

$$-4x^{2}-5x+9=\left(-4x-9\right)\left(x-1\right)$$

Evaluate

$$\left(1-x\right)\left(4x+9\right)$$