Solve 12-36k+15k^2

$$12 - 36 k + 15 k ^ { 2 }$$

Factor

$$3\left(k-2\right)\left(5k-2\right)$$

Solution Steps

Factor out 3.

$$3\left(4-12k+5k^{2}\right)$$

Consider 4-12k+5k^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$5k^{2}-12k+4$$

Factor the expression by grouping. First, the expression needs to be rewritten as 5k^{2}+ak+bk+4. To find a and b, set up a system to be solved.

$$a+b=-12 ab=5\times 4=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -12.

$$a=-10 b=-2$$

Rewrite 5k^{2}-12k+4 as \left(5k^{2}-10k\right)+\left(-2k+4\right).

$$\left(5k^{2}-10k\right)+\left(-2k+4\right)$$

Factor out 5k in the first and -2 in the second group.

$$5k\left(k-2\right)-2\left(k-2\right)$$

Factor out common term k-2 by using distributive property.

$$\left(k-2\right)\left(5k-2\right)$$

Rewrite the complete factored expression.

$$3\left(k-2\right)\left(5k-2\right)$$

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.

$$15k^{2}-36k+12=0$$

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.

$$k=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 15\times 12}}{2\times 15}$$

Square -36.

$$k=\frac{-\left(-36\right)±\sqrt{1296-4\times 15\times 12}}{2\times 15}$$

Multiply -4 times 15.

$$k=\frac{-\left(-36\right)±\sqrt{1296-60\times 12}}{2\times 15}$$

Multiply -60 times 12.

$$k=\frac{-\left(-36\right)±\sqrt{1296-720}}{2\times 15}$$

Add 1296 to -720.

$$k=\frac{-\left(-36\right)±\sqrt{576}}{2\times 15}$$

Take the square root of 576.

$$k=\frac{-\left(-36\right)±24}{2\times 15}$$

The opposite of -36 is 36.

$$k=\frac{36±24}{2\times 15}$$

Multiply 2 times 15.

$$k=\frac{36±24}{30}$$

Now solve the equation k=\frac{36±24}{30} when ± is plus. Add 36 to 24.

$$k=\frac{60}{30}$$

Divide 60 by 30.

$$k=2$$

Now solve the equation k=\frac{36±24}{30} when ± is minus. Subtract 24 from 36.

$$k=\frac{12}{30}$$

Reduce the fraction \frac{12}{30} to lowest terms by extracting and canceling out 6.

$$k=\frac{2}{5}$$

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

$$15k^{2}-36k+12=15\left(k-2\right)\left(k-\frac{2}{5}\right)$$

Subtract \frac{2}{5} from k by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$15k^{2}-36k+12=15\left(k-2\right)\times \frac{5k-2}{5}$$

Cancel out 5, the greatest common factor in 15 and 5.

$$15k^{2}-36k+12=3\left(k-2\right)\left(5k-2\right)$$

Evaluate

$$3\left(k-2\right)\left(5k-2\right)$$