Solve 9x^2+43x+28

$$9 { x }^{ 2 } +43x+28$$

Factor

$$\left(x+4\right)\left(9x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx+28. To find a and b, set up a system to be solved.

$$a+b=43 ab=9\times 28=252$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 252.

$$1,252 2,126 3,84 4,63 6,42 7,36 9,28 12,21 14,18$$

Calculate the sum for each pair.

$$1+252=253 2+126=128 3+84=87 4+63=67 6+42=48 7+36=43 9+28=37 12+21=33 14+18=32$$

The solution is the pair that gives sum 43.

$$a=7 b=36$$

Rewrite 9x^{2}+43x+28 as \left(9x^{2}+7x\right)+\left(36x+28\right).

$$\left(9x^{2}+7x\right)+\left(36x+28\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(9x+7\right)+4\left(9x+7\right)$$

Factor out common term 9x+7 by using distributive property.

$$\left(9x+7\right)\left(x+4\right)$$

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.

$$9x^{2}+43x+28=0$$

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.

$$x=\frac{-43±\sqrt{43^{2}-4\times 9\times 28}}{2\times 9}$$

Square 43.

$$x=\frac{-43±\sqrt{1849-4\times 9\times 28}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-43±\sqrt{1849-36\times 28}}{2\times 9}$$

Multiply -36 times 28.

$$x=\frac{-43±\sqrt{1849-1008}}{2\times 9}$$

Add 1849 to -1008.

$$x=\frac{-43±\sqrt{841}}{2\times 9}$$

Take the square root of 841.

$$x=\frac{-43±29}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-43±29}{18}$$

Now solve the equation x=\frac{-43±29}{18} when ± is plus. Add -43 to 29.

$$x=-\frac{14}{18}$$

Reduce the fraction \frac{-14}{18} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{9}$$

Now solve the equation x=\frac{-43±29}{18} when ± is minus. Subtract 29 from -43.

$$x=-\frac{72}{18}$$

Divide -72 by 18.

$$x=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{7}{9} for x_{1} and -4 for x_{2}.

$$9x^{2}+43x+28=9\left(x-\left(-\frac{7}{9}\right)\right)\left(x-\left(-4\right)\right)$$

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

$$9x^{2}+43x+28=9\left(x+\frac{7}{9}\right)\left(x+4\right)$$

Add \frac{7}{9} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9x^{2}+43x+28=9\times \frac{9x+7}{9}\left(x+4\right)$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9x^{2}+43x+28=\left(9x+7\right)\left(x+4\right)$$

Evaluate

$$\left(x+4\right)\left(9x+7\right)$$