Solve 3x^2+17x+10

$$3 { x }^{ 2 } +17x+10$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=17 ab=3\times 10=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 17.

$$a=2 b=15$$

Rewrite 3x^{2}+17x+10 as \left(3x^{2}+2x\right)+\left(15x+10\right).

$$\left(3x^{2}+2x\right)+\left(15x+10\right)$$

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

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

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

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

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.

$$3x^{2}+17x+10=0$$

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.

$$x=\frac{-17±\sqrt{17^{2}-4\times 3\times 10}}{2\times 3}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 3\times 10}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-17±\sqrt{289-12\times 10}}{2\times 3}$$

Multiply -12 times 10.

$$x=\frac{-17±\sqrt{289-120}}{2\times 3}$$

Add 289 to -120.

$$x=\frac{-17±\sqrt{169}}{2\times 3}$$

Take the square root of 169.

$$x=\frac{-17±13}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-17±13}{6}$$

Now solve the equation x=\frac{-17±13}{6} when ± is plus. Add -17 to 13.

$$x=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-17±13}{6} when ± is minus. Subtract 13 from -17.

$$x=-\frac{30}{6}$$

Divide -30 by 6.

$$x=-5$$

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

$$3x^{2}+17x+10=3\left(x-\left(-\frac{2}{3}\right)\right)\left(x-\left(-5\right)\right)$$

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

$$3x^{2}+17x+10=3\left(x+\frac{2}{3}\right)\left(x+5\right)$$

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

$$3x^{2}+17x+10=3\times \frac{3x+2}{3}\left(x+5\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}+17x+10=\left(3x+2\right)\left(x+5\right)$$

Evaluate

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