Solve 13x-10+3x^2

$$13 x - 10 + 3 x ^ { 2 }$$

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3x^{2}+13x-10$$

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.

$$a+b=13 ab=3\left(-10\right)=-30$$

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 -30.

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

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 13.

$$a=-2 b=15$$

Rewrite 3x^{2}+13x-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.

$$3x^{2}+13x-10=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\times 3\left(-10\right)}}{2\times 3}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 3\left(-10\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-13±\sqrt{169-12\left(-10\right)}}{2\times 3}$$

Multiply -12 times -10.

$$x=\frac{-13±\sqrt{169+120}}{2\times 3}$$

Add 169 to 120.

$$x=\frac{-13±\sqrt{289}}{2\times 3}$$

Take the square root of 289.

$$x=\frac{-13±17}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-13±17}{6}$$

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

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

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{-13±17}{6} when ± is minus. Subtract 17 from -13.

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

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}.

$$3x^{2}+13x-10=3\left(x-\frac{2}{3}\right)\left(x-\left(-5\right)\right)$$

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

$$3x^{2}+13x-10=3\left(x-\frac{2}{3}\right)\left(x+5\right)$$

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

$$3x^{2}+13x-10=3\times \frac{3x-2}{3}\left(x+5\right)$$

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

$$3x^{2}+13x-10=\left(3x-2\right)\left(x+5\right)$$

Evaluate

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