Solve 50+35x+6x^2

$$50 + 35 x + 6 x ^ { 2 }$$

Factor

$$\left(2x+5\right)\left(3x+10\right)$$

Solution Steps

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

$$6x^{2}+35x+50$$

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+50. To find a and b, set up a system to be solved.

$$a+b=35 ab=6\times 50=300$$

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 300.

$$1,300 2,150 3,100 4,75 5,60 6,50 10,30 12,25 15,20$$

Calculate the sum for each pair.

$$1+300=301 2+150=152 3+100=103 4+75=79 5+60=65 6+50=56 10+30=40 12+25=37 15+20=35$$

The solution is the pair that gives sum 35.

$$a=15 b=20$$

Rewrite 6x^{2}+35x+50 as \left(6x^{2}+15x\right)+\left(20x+50\right).

$$\left(6x^{2}+15x\right)+\left(20x+50\right)$$

Factor out 3x in the first and 10 in the second group.

$$3x\left(2x+5\right)+10\left(2x+5\right)$$

Factor out common term 2x+5 by using distributive property.

$$\left(2x+5\right)\left(3x+10\right)$$

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.

$$6x^{2}+35x+50=0$$

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.

$$x=\frac{-35±\sqrt{35^{2}-4\times 6\times 50}}{2\times 6}$$

Square 35.

$$x=\frac{-35±\sqrt{1225-4\times 6\times 50}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-35±\sqrt{1225-24\times 50}}{2\times 6}$$

Multiply -24 times 50.

$$x=\frac{-35±\sqrt{1225-1200}}{2\times 6}$$

Add 1225 to -1200.

$$x=\frac{-35±\sqrt{25}}{2\times 6}$$

Take the square root of 25.

$$x=\frac{-35±5}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-35±5}{12}$$

Now solve the equation x=\frac{-35±5}{12} when ± is plus. Add -35 to 5.

$$x=-\frac{30}{12}$$

Reduce the fraction \frac{-30}{12} to lowest terms by extracting and canceling out 6.

$$x=-\frac{5}{2}$$

Now solve the equation x=\frac{-35±5}{12} when ± is minus. Subtract 5 from -35.

$$x=-\frac{40}{12}$$

Reduce the fraction \frac{-40}{12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{10}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{2} for x_{1} and -\frac{10}{3} for x_{2}.

$$6x^{2}+35x+50=6\left(x-\left(-\frac{5}{2}\right)\right)\left(x-\left(-\frac{10}{3}\right)\right)$$

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

$$6x^{2}+35x+50=6\left(x+\frac{5}{2}\right)\left(x+\frac{10}{3}\right)$$

Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+35x+50=6\times \frac{2x+5}{2}\left(x+\frac{10}{3}\right)$$

Add \frac{10}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+35x+50=6\times \frac{2x+5}{2}\times \frac{3x+10}{3}$$

Multiply \frac{2x+5}{2} times \frac{3x+10}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+35x+50=6\times \frac{\left(2x+5\right)\left(3x+10\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}+35x+50=6\times \frac{\left(2x+5\right)\left(3x+10\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6x^{2}+35x+50=\left(2x+5\right)\left(3x+10\right)$$

Evaluate

$$\left(2x+5\right)\left(3x+10\right)$$