Solve 10x^2+43x-9

$$10 { x }^{ 2 } +43x-9$$

Factor

$$\left(5x-1\right)\left(2x+9\right)$$

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.

$$a+b=43 ab=10\left(-9\right)=-90$$

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 -90.

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

Calculate the sum for each pair.

$$-1+90=89 -2+45=43 -3+30=27 -5+18=13 -6+15=9 -9+10=1$$

The solution is the pair that gives sum 43.

$$a=-2 b=45$$

Rewrite 10x^{2}+43x-9 as \left(10x^{2}-2x\right)+\left(45x-9\right).

$$\left(10x^{2}-2x\right)+\left(45x-9\right)$$

Factor out 2x in the first and 9 in the second group.

$$2x\left(5x-1\right)+9\left(5x-1\right)$$

Factor out common term 5x-1 by using distributive property.

$$\left(5x-1\right)\left(2x+9\right)$$

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.

$$10x^{2}+43x-9=0$$

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.

$$x=\frac{-43±\sqrt{43^{2}-4\times 10\left(-9\right)}}{2\times 10}$$

Square 43.

$$x=\frac{-43±\sqrt{1849-4\times 10\left(-9\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-43±\sqrt{1849-40\left(-9\right)}}{2\times 10}$$

Multiply -40 times -9.

$$x=\frac{-43±\sqrt{1849+360}}{2\times 10}$$

Add 1849 to 360.

$$x=\frac{-43±\sqrt{2209}}{2\times 10}$$

Take the square root of 2209.

$$x=\frac{-43±47}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-43±47}{20}$$

Now solve the equation x=\frac{-43±47}{20} when ± is plus. Add -43 to 47.

$$x=\frac{4}{20}$$

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

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-43±47}{20} when ± is minus. Subtract 47 from -43.

$$x=-\frac{90}{20}$$

Reduce the fraction \frac{-90}{20} to lowest terms by extracting and canceling out 10.

$$x=-\frac{9}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{5} for x_{1} and -\frac{9}{2} for x_{2}.

$$10x^{2}+43x-9=10\left(x-\frac{1}{5}\right)\left(x-\left(-\frac{9}{2}\right)\right)$$

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

$$10x^{2}+43x-9=10\left(x-\frac{1}{5}\right)\left(x+\frac{9}{2}\right)$$

Subtract \frac{1}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+43x-9=10\times \frac{5x-1}{5}\left(x+\frac{9}{2}\right)$$

Add \frac{9}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+43x-9=10\times \frac{5x-1}{5}\times \frac{2x+9}{2}$$

Multiply \frac{5x-1}{5} times \frac{2x+9}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+43x-9=10\times \frac{\left(5x-1\right)\left(2x+9\right)}{5\times 2}$$

Multiply 5 times 2.

$$10x^{2}+43x-9=10\times \frac{\left(5x-1\right)\left(2x+9\right)}{10}$$

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

$$10x^{2}+43x-9=\left(5x-1\right)\left(2x+9\right)$$

Evaluate

$$\left(5x-1\right)\left(2x+9\right)$$