Solve 5x^2-41x+42

$$5 { x }^{ 2 } -41x+42$$

Factor

$$\left(x-7\right)\left(5x-6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+42. To find a and b, set up a system to be solved.

$$a+b=-41 ab=5\times 42=210$$

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 210.

$$-1,-210 -2,-105 -3,-70 -5,-42 -6,-35 -7,-30 -10,-21 -14,-15$$

Calculate the sum for each pair.

$$-1-210=-211 -2-105=-107 -3-70=-73 -5-42=-47 -6-35=-41 -7-30=-37 -10-21=-31 -14-15=-29$$

The solution is the pair that gives sum -41.

$$a=-35 b=-6$$

Rewrite 5x^{2}-41x+42 as \left(5x^{2}-35x\right)+\left(-6x+42\right).

$$\left(5x^{2}-35x\right)+\left(-6x+42\right)$$

Factor out 5x in the first and -6 in the second group.

$$5x\left(x-7\right)-6\left(x-7\right)$$

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

$$\left(x-7\right)\left(5x-6\right)$$

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.

$$5x^{2}-41x+42=0$$

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.

$$x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 5\times 42}}{2\times 5}$$

Square -41.

$$x=\frac{-\left(-41\right)±\sqrt{1681-4\times 5\times 42}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-41\right)±\sqrt{1681-20\times 42}}{2\times 5}$$

Multiply -20 times 42.

$$x=\frac{-\left(-41\right)±\sqrt{1681-840}}{2\times 5}$$

Add 1681 to -840.

$$x=\frac{-\left(-41\right)±\sqrt{841}}{2\times 5}$$

Take the square root of 841.

$$x=\frac{-\left(-41\right)±29}{2\times 5}$$

The opposite of -41 is 41.

$$x=\frac{41±29}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{41±29}{10}$$

Now solve the equation x=\frac{41±29}{10} when ± is plus. Add 41 to 29.

$$x=\frac{70}{10}$$

Divide 70 by 10.

$$x=7$$

Now solve the equation x=\frac{41±29}{10} when ± is minus. Subtract 29 from 41.

$$x=\frac{12}{10}$$

Reduce the fraction \frac{12}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{6}{5}$$

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 \frac{6}{5} for x_{2}.

$$5x^{2}-41x+42=5\left(x-7\right)\left(x-\frac{6}{5}\right)$$

Subtract \frac{6}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}-41x+42=5\left(x-7\right)\times \frac{5x-6}{5}$$

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

$$5x^{2}-41x+42=\left(x-7\right)\left(5x-6\right)$$

Evaluate

$$\left(x-7\right)\left(5x-6\right)$$