Solve v^2+36v+35

$$v ^ { 2 } + 36 v + 35$$

Factor

$$\left(v+1\right)\left(v+35\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as v^{2}+av+bv+35. To find a and b, set up a system to be solved.

$$a+b=36 ab=1\times 35=35$$

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 35.

$$1,35 5,7$$

Calculate the sum for each pair.

$$1+35=36 5+7=12$$

The solution is the pair that gives sum 36.

$$a=1 b=35$$

Rewrite v^{2}+36v+35 as \left(v^{2}+v\right)+\left(35v+35\right).

$$\left(v^{2}+v\right)+\left(35v+35\right)$$

Factor out v in the first and 35 in the second group.

$$v\left(v+1\right)+35\left(v+1\right)$$

Factor out common term v+1 by using distributive property.

$$\left(v+1\right)\left(v+35\right)$$

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.

$$v^{2}+36v+35=0$$

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.

$$v=\frac{-36±\sqrt{36^{2}-4\times 35}}{2}$$

Square 36.

$$v=\frac{-36±\sqrt{1296-4\times 35}}{2}$$

Multiply -4 times 35.

$$v=\frac{-36±\sqrt{1296-140}}{2}$$

Add 1296 to -140.

$$v=\frac{-36±\sqrt{1156}}{2}$$

Take the square root of 1156.

$$v=\frac{-36±34}{2}$$

Now solve the equation v=\frac{-36±34}{2} when ± is plus. Add -36 to 34.

$$v=-\frac{2}{2}$$

Divide -2 by 2.

$$v=-1$$

Now solve the equation v=\frac{-36±34}{2} when ± is minus. Subtract 34 from -36.

$$v=-\frac{70}{2}$$

Divide -70 by 2.

$$v=-35$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and -35 for x_{2}.

$$v^{2}+36v+35=\left(v-\left(-1\right)\right)\left(v-\left(-35\right)\right)$$

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

$$v^{2}+36v+35=\left(v+1\right)\left(v+35\right)$$

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.

$$x ^ 2 +36x +35 = 0$$

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

$$r + s = -36 rs = 35$$

Two numbers r and s sum up to -36 exactly when the average of the two numbers is \frac{1}{2}*-36 = -18. 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>

$$r = -18 - u s = -18 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 35

$$(-18 - u) (-18 + u) = 35$$

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

$$324 - u^2 = 35$$

Simplify the expression by subtracting 324 on both sides

$$-u^2 = 35-324 = -289$$

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

$$u^2 = 289 u = \pm\sqrt{289} = \pm 17$$

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

$$r =-18 - 17 = -35 s = -18 + 17 = -1$$

Evaluate

$$\left(v+1\right)\left(v+35\right)$$