Solve j^2-13j+42

$${ j }^{ 2 } -13j+42$$

Factor

$$\left(j-7\right)\left(j-6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as j^{2}+aj+bj+42. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 42=42$$

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 42.

$$-1,-42 -2,-21 -3,-14 -6,-7$$

Calculate the sum for each pair.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

The solution is the pair that gives sum -13.

$$a=-7 b=-6$$

Rewrite j^{2}-13j+42 as \left(j^{2}-7j\right)+\left(-6j+42\right).

$$\left(j^{2}-7j\right)+\left(-6j+42\right)$$

Factor out j in the first and -6 in the second group.

$$j\left(j-7\right)-6\left(j-7\right)$$

Factor out common term j-7 by using distributive property.

$$\left(j-7\right)\left(j-6\right)$$

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.

$$j^{2}-13j+42=0$$

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.

$$j=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 42}}{2}$$

Square -13.

$$j=\frac{-\left(-13\right)±\sqrt{169-4\times 42}}{2}$$

Multiply -4 times 42.

$$j=\frac{-\left(-13\right)±\sqrt{169-168}}{2}$$

Add 169 to -168.

$$j=\frac{-\left(-13\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$j=\frac{-\left(-13\right)±1}{2}$$

The opposite of -13 is 13.

$$j=\frac{13±1}{2}$$

Now solve the equation j=\frac{13±1}{2} when ± is plus. Add 13 to 1.

$$j=\frac{14}{2}$$

Divide 14 by 2.

$$j=7$$

Now solve the equation j=\frac{13±1}{2} when ± is minus. Subtract 1 from 13.

$$j=\frac{12}{2}$$

Divide 12 by 2.

$$j=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 7 for x_{1} and 6 for x_{2}.

$$j^{2}-13j+42=\left(j-7\right)\left(j-6\right)$$

Evaluate

$$\left(j-7\right)\left(j-6\right)$$