Solve 35v_1^2-47v_{1}+6

$$35 v _ { 1 } ^ { 2 } - 47 v _ { 1 } + 6$$

Factor

$$\left(5v_{1}-6\right)\left(7v_{1}-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 35v_{1}^{2}+av_{1}+bv_{1}+6. To find a and b, set up a system to be solved.

$$a+b=-47 ab=35\times 6=210$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 210.

$$-1,-210 -2,-105 -3,-70 -5,-42 -6,-35 -7,-30 -10,-21 -14,-15$$

Calculate the sum for each pair.

$$-1-210=-211 -2-105=-107 -3-70=-73 -5-42=-47 -6-35=-41 -7-30=-37 -10-21=-31 -14-15=-29$$

The solution is the pair that gives sum -47.

$$a=-42 b=-5$$

Rewrite 35v_{1}^{2}-47v_{1}+6 as \left(35v_{1}^{2}-42v_{1}\right)+\left(-5v_{1}+6\right).

$$\left(35v_{1}^{2}-42v_{1}\right)+\left(-5v_{1}+6\right)$$

Factor out 7v_{1} in the first and -1 in the second group.

$$7v_{1}\left(5v_{1}-6\right)-\left(5v_{1}-6\right)$$

Factor out common term 5v_{1}-6 by using distributive property.

$$\left(5v_{1}-6\right)\left(7v_{1}-1\right)$$

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.

$$35v_{1}^{2}-47v_{1}+6=0$$

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.

$$v_{1}=\frac{-\left(-47\right)±\sqrt{\left(-47\right)^{2}-4\times 35\times 6}}{2\times 35}$$

Square -47.

$$v_{1}=\frac{-\left(-47\right)±\sqrt{2209-4\times 35\times 6}}{2\times 35}$$

Multiply -4 times 35.

$$v_{1}=\frac{-\left(-47\right)±\sqrt{2209-140\times 6}}{2\times 35}$$

Multiply -140 times 6.

$$v_{1}=\frac{-\left(-47\right)±\sqrt{2209-840}}{2\times 35}$$

Add 2209 to -840.

$$v_{1}=\frac{-\left(-47\right)±\sqrt{1369}}{2\times 35}$$

Take the square root of 1369.

$$v_{1}=\frac{-\left(-47\right)±37}{2\times 35}$$

The opposite of -47 is 47.

$$v_{1}=\frac{47±37}{2\times 35}$$

Multiply 2 times 35.

$$v_{1}=\frac{47±37}{70}$$

Now solve the equation v_{1}=\frac{47±37}{70} when ± is plus. Add 47 to 37.

$$v_{1}=\frac{84}{70}$$

Reduce the fraction \frac{84}{70} to lowest terms by extracting and canceling out 14.

$$v_{1}=\frac{6}{5}$$

Now solve the equation v_{1}=\frac{47±37}{70} when ± is minus. Subtract 37 from 47.

$$v_{1}=\frac{10}{70}$$

Reduce the fraction \frac{10}{70} to lowest terms by extracting and canceling out 10.

$$v_{1}=\frac{1}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{6}{5} for x_{1} and \frac{1}{7} for x_{2}.

$$35v_{1}^{2}-47v_{1}+6=35\left(v_{1}-\frac{6}{5}\right)\left(v_{1}-\frac{1}{7}\right)$$

Subtract \frac{6}{5} from v_{1} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$35v_{1}^{2}-47v_{1}+6=35\times \frac{5v_{1}-6}{5}\left(v_{1}-\frac{1}{7}\right)$$

Subtract \frac{1}{7} from v_{1} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$35v_{1}^{2}-47v_{1}+6=35\times \frac{5v_{1}-6}{5}\times \frac{7v_{1}-1}{7}$$

Multiply \frac{5v_{1}-6}{5} times \frac{7v_{1}-1}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$35v_{1}^{2}-47v_{1}+6=35\times \frac{\left(5v_{1}-6\right)\left(7v_{1}-1\right)}{5\times 7}$$

Multiply 5 times 7.

$$35v_{1}^{2}-47v_{1}+6=35\times \frac{\left(5v_{1}-6\right)\left(7v_{1}-1\right)}{35}$$

Cancel out 35, the greatest common factor in 35 and 35.

$$35v_{1}^{2}-47v_{1}+6=\left(5v_{1}-6\right)\left(7v_{1}-1\right)$$

Evaluate

$$\left(5v_{1}-6\right)\left(7v_{1}-1\right)$$