Solve 10x^2-21x+9

$$10 x ^ { 2 } - 21 x + 9$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-21 ab=10\times 9=90$$

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 90.

$$-1,-90 -2,-45 -3,-30 -5,-18 -6,-15 -9,-10$$

Calculate the sum for each pair.

$$-1-90=-91 -2-45=-47 -3-30=-33 -5-18=-23 -6-15=-21 -9-10=-19$$

The solution is the pair that gives sum -21.

$$a=-15 b=-6$$

Rewrite 10x^{2}-21x+9 as \left(10x^{2}-15x\right)+\left(-6x+9\right).

$$\left(10x^{2}-15x\right)+\left(-6x+9\right)$$

Factor out 5x in the first and -3 in the second group.

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

Factor out common term 2x-3 by using distributive property.

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

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.

$$10x^{2}-21x+9=0$$

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.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 10\times 9}}{2\times 10}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 10\times 9}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-21\right)±\sqrt{441-40\times 9}}{2\times 10}$$

Multiply -40 times 9.

$$x=\frac{-\left(-21\right)±\sqrt{441-360}}{2\times 10}$$

Add 441 to -360.

$$x=\frac{-\left(-21\right)±\sqrt{81}}{2\times 10}$$

Take the square root of 81.

$$x=\frac{-\left(-21\right)±9}{2\times 10}$$

The opposite of -21 is 21.

$$x=\frac{21±9}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{21±9}{20}$$

Now solve the equation x=\frac{21±9}{20} when ± is plus. Add 21 to 9.

$$x=\frac{30}{20}$$

Reduce the fraction \frac{30}{20} to lowest terms by extracting and canceling out 10.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{21±9}{20} when ± is minus. Subtract 9 from 21.

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

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

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

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

$$10x^{2}-21x+9=10\left(x-\frac{3}{2}\right)\left(x-\frac{3}{5}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}-21x+9=10\times \frac{2x-3}{2}\left(x-\frac{3}{5}\right)$$

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

$$10x^{2}-21x+9=10\times \frac{2x-3}{2}\times \frac{5x-3}{5}$$

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

$$10x^{2}-21x+9=10\times \frac{\left(2x-3\right)\left(5x-3\right)}{2\times 5}$$

Multiply 2 times 5.

$$10x^{2}-21x+9=10\times \frac{\left(2x-3\right)\left(5x-3\right)}{10}$$

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

$$10x^{2}-21x+9=\left(2x-3\right)\left(5x-3\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 10

$$x ^ 2 -\frac{21}{10}x +\frac{9}{10} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{21}{10} rs = \frac{9}{10}$$

Two numbers r and s sum up to \frac{21}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{21}{10} = \frac{21}{20}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{21}{20} - u s = \frac{21}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{10}

$$(\frac{21}{20} - u) (\frac{21}{20} + u) = \frac{9}{10}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{441}{400} - u^2 = \frac{9}{10}$$

Simplify the expression by subtracting \frac{441}{400} on both sides

$$-u^2 = \frac{9}{10}-\frac{441}{400} = -\frac{81}{400}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{81}{400} u = \pm\sqrt{\frac{81}{400}} = \pm \frac{9}{20}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{21}{20} - \frac{9}{20} = 0.600 s = \frac{21}{20} + \frac{9}{20} = 1.500$$

Evaluate

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