Solve 10x^2+29x-21

$$10 { x }^{ 2 } +29x-21$$

Factor

$$\left(5x-3\right)\left(2x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=29 ab=10\left(-21\right)=-210$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the 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.

$$-1+210=209 -2+105=103 -3+70=67 -5+42=37 -6+35=29 -7+30=23 -10+21=11 -14+15=1$$

The solution is the pair that gives sum 29.

$$a=-6 b=35$$

Rewrite 10x^{2}+29x-21 as \left(10x^{2}-6x\right)+\left(35x-21\right).

$$\left(10x^{2}-6x\right)+\left(35x-21\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(5x-3\right)+7\left(5x-3\right)$$

Factor out common term 5x-3 by using distributive property.

$$\left(5x-3\right)\left(2x+7\right)$$

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.

$$10x^{2}+29x-21=0$$

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.

$$x=\frac{-29±\sqrt{29^{2}-4\times 10\left(-21\right)}}{2\times 10}$$

Square 29.

$$x=\frac{-29±\sqrt{841-4\times 10\left(-21\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-29±\sqrt{841-40\left(-21\right)}}{2\times 10}$$

Multiply -40 times -21.

$$x=\frac{-29±\sqrt{841+840}}{2\times 10}$$

Add 841 to 840.

$$x=\frac{-29±\sqrt{1681}}{2\times 10}$$

Take the square root of 1681.

$$x=\frac{-29±41}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-29±41}{20}$$

Now solve the equation x=\frac{-29±41}{20} when ± is plus. Add -29 to 41.

$$x=\frac{12}{20}$$

Reduce the fraction \frac{12}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{5}$$

Now solve the equation x=\frac{-29±41}{20} when ± is minus. Subtract 41 from -29.

$$x=-\frac{70}{20}$$

Reduce the fraction \frac{-70}{20} to lowest terms by extracting and canceling out 10.

$$x=-\frac{7}{2}$$

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

$$10x^{2}+29x-21=10\left(x-\frac{3}{5}\right)\left(x-\left(-\frac{7}{2}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$10x^{2}+29x-21=10\left(x-\frac{3}{5}\right)\left(x+\frac{7}{2}\right)$$

Subtract \frac{3}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+29x-21=10\times \frac{5x-3}{5}\left(x+\frac{7}{2}\right)$$

Add \frac{7}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+29x-21=10\times \frac{5x-3}{5}\times \frac{2x+7}{2}$$

Multiply \frac{5x-3}{5} times \frac{2x+7}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+29x-21=10\times \frac{\left(5x-3\right)\left(2x+7\right)}{5\times 2}$$

Multiply 5 times 2.

$$10x^{2}+29x-21=10\times \frac{\left(5x-3\right)\left(2x+7\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10x^{2}+29x-21=\left(5x-3\right)\left(2x+7\right)$$

Evaluate

$$\left(5x-3\right)\left(2x+7\right)$$