Solve 40x^2-71x+21

$$40 { x }^{ 2 } -71x+21$$

Factor

$$\left(5x-7\right)\left(8x-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 40x^{2}+ax+bx+21. To find a and b, set up a system to be solved.

$$a+b=-71 ab=40\times 21=840$$

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 840.

$$-1,-840 -2,-420 -3,-280 -4,-210 -5,-168 -6,-140 -7,-120 -8,-105 -10,-84 -12,-70 -14,-60 -15,-56 -20,-42 -21,-40 -24,-35 -28,-30$$

Calculate the sum for each pair.

$$-1-840=-841 -2-420=-422 -3-280=-283 -4-210=-214 -5-168=-173 -6-140=-146 -7-120=-127 -8-105=-113 -10-84=-94 -12-70=-82 -14-60=-74 -15-56=-71 -20-42=-62 -21-40=-61 -24-35=-59 -28-30=-58$$

The solution is the pair that gives sum -71.

$$a=-56 b=-15$$

Rewrite 40x^{2}-71x+21 as \left(40x^{2}-56x\right)+\left(-15x+21\right).

$$\left(40x^{2}-56x\right)+\left(-15x+21\right)$$

Factor out 8x in the first and -3 in the second group.

$$8x\left(5x-7\right)-3\left(5x-7\right)$$

Factor out common term 5x-7 by using distributive property.

$$\left(5x-7\right)\left(8x-3\right)$$

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.

$$40x^{2}-71x+21=0$$

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.

$$x=\frac{-\left(-71\right)±\sqrt{\left(-71\right)^{2}-4\times 40\times 21}}{2\times 40}$$

Square -71.

$$x=\frac{-\left(-71\right)±\sqrt{5041-4\times 40\times 21}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-\left(-71\right)±\sqrt{5041-160\times 21}}{2\times 40}$$

Multiply -160 times 21.

$$x=\frac{-\left(-71\right)±\sqrt{5041-3360}}{2\times 40}$$

Add 5041 to -3360.

$$x=\frac{-\left(-71\right)±\sqrt{1681}}{2\times 40}$$

Take the square root of 1681.

$$x=\frac{-\left(-71\right)±41}{2\times 40}$$

The opposite of -71 is 71.

$$x=\frac{71±41}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{71±41}{80}$$

Now solve the equation x=\frac{71±41}{80} when ± is plus. Add 71 to 41.

$$x=\frac{112}{80}$$

Reduce the fraction \frac{112}{80} to lowest terms by extracting and canceling out 16.

$$x=\frac{7}{5}$$

Now solve the equation x=\frac{71±41}{80} when ± is minus. Subtract 41 from 71.

$$x=\frac{30}{80}$$

Reduce the fraction \frac{30}{80} to lowest terms by extracting and canceling out 10.

$$x=\frac{3}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{5} for x_{1} and \frac{3}{8} for x_{2}.

$$40x^{2}-71x+21=40\left(x-\frac{7}{5}\right)\left(x-\frac{3}{8}\right)$$

Subtract \frac{7}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$40x^{2}-71x+21=40\times \frac{5x-7}{5}\left(x-\frac{3}{8}\right)$$

Subtract \frac{3}{8} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$40x^{2}-71x+21=40\times \frac{5x-7}{5}\times \frac{8x-3}{8}$$

Multiply \frac{5x-7}{5} times \frac{8x-3}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$40x^{2}-71x+21=40\times \frac{\left(5x-7\right)\left(8x-3\right)}{5\times 8}$$

Multiply 5 times 8.

$$40x^{2}-71x+21=40\times \frac{\left(5x-7\right)\left(8x-3\right)}{40}$$

Cancel out 40, the greatest common factor in 40 and 40.

$$40x^{2}-71x+21=\left(5x-7\right)\left(8x-3\right)$$

Evaluate

$$\left(5x-7\right)\left(8x-3\right)$$