Solve 4x^2-15x+9

$$4 { x }^{ 2 } -15x+9$$

Factor

$$\left(x-3\right)\left(4x-3\right)$$

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.

$$a+b=-15 ab=4\times 9=36$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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 -15.

$$a=-12 b=-3$$

Rewrite 4x^{2}-15x+9 as \left(4x^{2}-12x\right)+\left(-3x+9\right).

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

Factor out 4x in the first and -3 in the second group.

$$4x\left(x-3\right)-3\left(x-3\right)$$

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

$$\left(x-3\right)\left(4x-3\right)$$

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.

$$4x^{2}-15x+9=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 4\times 9}}{2\times 4}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 4\times 9}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-15\right)±\sqrt{225-16\times 9}}{2\times 4}$$

Multiply -16 times 9.

$$x=\frac{-\left(-15\right)±\sqrt{225-144}}{2\times 4}$$

Add 225 to -144.

$$x=\frac{-\left(-15\right)±\sqrt{81}}{2\times 4}$$

Take the square root of 81.

$$x=\frac{-\left(-15\right)±9}{2\times 4}$$

The opposite of -15 is 15.

$$x=\frac{15±9}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{15±9}{8}$$

Now solve the equation x=\frac{15±9}{8} when ± is plus. Add 15 to 9.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{15±9}{8} when ± is minus. Subtract 9 from 15.

$$x=\frac{6}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and \frac{3}{4} for x_{2}.

$$4x^{2}-15x+9=4\left(x-3\right)\left(x-\frac{3}{4}\right)$$

Subtract \frac{3}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-15x+9=4\left(x-3\right)\times \frac{4x-3}{4}$$

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

$$4x^{2}-15x+9=\left(x-3\right)\left(4x-3\right)$$

Evaluate

$$\left(x-3\right)\left(4x-3\right)$$