Solve f(x)=12x-9-4x^2

$$f ( x ) = 12 x - 9 - 4 x ^ { 2 }$$

Factor

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

Solution Steps

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

$$-4x^{2}+12x-9$$

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.

$$a+b=12 ab=-4\left(-9\right)=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both 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.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 12.

$$a=6 b=6$$

Rewrite -4x^{2}+12x-9 as \left(-4x^{2}+6x\right)+\left(6x-9\right).

$$\left(-4x^{2}+6x\right)+\left(6x-9\right)$$

Factor out -2x in the first and 3 in the second group.

$$-2x\left(2x-3\right)+3\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(-2x+3\right)$$

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.

$$-4x^{2}+12x-9=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-4\right)\left(-9\right)}}{2\left(-4\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-4\right)\left(-9\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-12±\sqrt{144+16\left(-9\right)}}{2\left(-4\right)}$$

Multiply 16 times -9.

$$x=\frac{-12±\sqrt{144-144}}{2\left(-4\right)}$$

Add 144 to -144.

$$x=\frac{-12±\sqrt{0}}{2\left(-4\right)}$$

Take the square root of 0.

$$x=\frac{-12±0}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-12±0}{-8}$$

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

$$-4x^{2}+12x-9=-4\left(x-\frac{3}{2}\right)\left(x-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+12x-9=-4\times \frac{-2x+3}{-2}\left(x-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+12x-9=-4\times \frac{-2x+3}{-2}\times \frac{-2x+3}{-2}$$

Multiply \frac{-2x+3}{-2} times \frac{-2x+3}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+12x-9=-4\times \frac{\left(-2x+3\right)\left(-2x+3\right)}{-2\left(-2\right)}$$

Multiply -2 times -2.

$$-4x^{2}+12x-9=-4\times \frac{\left(-2x+3\right)\left(-2x+3\right)}{4}$$

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

$$-4x^{2}+12x-9=-\left(-2x+3\right)\left(-2x+3\right)$$

Evaluate

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