Solve 4q^2-34q+70

$$4 q ^ { 2 } - 34 q + 70$$

Factor

$$2\left(q-5\right)\left(2q-7\right)$$

Solution Steps

Factor out 2.

$$2\left(2q^{2}-17q+35\right)$$

Consider 2q^{2}-17q+35. Factor the expression by grouping. First, the expression needs to be rewritten as 2q^{2}+aq+bq+35. To find a and b, set up a system to be solved.

$$a+b=-17 ab=2\times 35=70$$

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 70.

$$-1,-70 -2,-35 -5,-14 -7,-10$$

Calculate the sum for each pair.

$$-1-70=-71 -2-35=-37 -5-14=-19 -7-10=-17$$

The solution is the pair that gives sum -17.

$$a=-10 b=-7$$

Rewrite 2q^{2}-17q+35 as \left(2q^{2}-10q\right)+\left(-7q+35\right).

$$\left(2q^{2}-10q\right)+\left(-7q+35\right)$$

Factor out 2q in the first and -7 in the second group.

$$2q\left(q-5\right)-7\left(q-5\right)$$

Factor out common term q-5 by using distributive property.

$$\left(q-5\right)\left(2q-7\right)$$

Rewrite the complete factored expression.

$$2\left(q-5\right)\left(2q-7\right)$$

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.

$$4q^{2}-34q+70=0$$

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.

$$q=\frac{-\left(-34\right)±\sqrt{\left(-34\right)^{2}-4\times 4\times 70}}{2\times 4}$$

Square -34.

$$q=\frac{-\left(-34\right)±\sqrt{1156-4\times 4\times 70}}{2\times 4}$$

Multiply -4 times 4.

$$q=\frac{-\left(-34\right)±\sqrt{1156-16\times 70}}{2\times 4}$$

Multiply -16 times 70.

$$q=\frac{-\left(-34\right)±\sqrt{1156-1120}}{2\times 4}$$

Add 1156 to -1120.

$$q=\frac{-\left(-34\right)±\sqrt{36}}{2\times 4}$$

Take the square root of 36.

$$q=\frac{-\left(-34\right)±6}{2\times 4}$$

The opposite of -34 is 34.

$$q=\frac{34±6}{2\times 4}$$

Multiply 2 times 4.

$$q=\frac{34±6}{8}$$

Now solve the equation q=\frac{34±6}{8} when ± is plus. Add 34 to 6.

$$q=\frac{40}{8}$$

Divide 40 by 8.

$$q=5$$

Now solve the equation q=\frac{34±6}{8} when ± is minus. Subtract 6 from 34.

$$q=\frac{28}{8}$$

Reduce the fraction \frac{28}{8} to lowest terms by extracting and canceling out 4.

$$q=\frac{7}{2}$$

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

$$4q^{2}-34q+70=4\left(q-5\right)\left(q-\frac{7}{2}\right)$$

Subtract \frac{7}{2} from q by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4q^{2}-34q+70=4\left(q-5\right)\times \frac{2q-7}{2}$$

Cancel out 2, the greatest common factor in 4 and 2.

$$4q^{2}-34q+70=2\left(q-5\right)\left(2q-7\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 4

$$x ^ 2 -\frac{17}{2}x +\frac{35}{2} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{17}{2} rs = \frac{35}{2}$$

Two numbers r and s sum up to \frac{17}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{2} = \frac{17}{4}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{17}{4} - u s = \frac{17}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{35}{2}

$$(\frac{17}{4} - u) (\frac{17}{4} + u) = \frac{35}{2}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{289}{16} - u^2 = \frac{35}{2}$$

Simplify the expression by subtracting \frac{289}{16} on both sides

$$-u^2 = \frac{35}{2}-\frac{289}{16} = -\frac{9}{16}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{9}{16} u = \pm\sqrt{\frac{9}{16}} = \pm \frac{3}{4}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{17}{4} - \frac{3}{4} = 3.500 s = \frac{17}{4} + \frac{3}{4} = 5$$

Evaluate

$$2\left(q-5\right)\left(2q-7\right)$$