Solve z^2+34z+64

$$z ^ { 2 } + 34 z + 64$$

Factor

$$\left(z+2\right)\left(z+32\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz+64. To find a and b, set up a system to be solved.

$$a+b=34 ab=1\times 64=64$$

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 64.

$$1,64 2,32 4,16 8,8$$

Calculate the sum for each pair.

$$1+64=65 2+32=34 4+16=20 8+8=16$$

The solution is the pair that gives sum 34.

$$a=2 b=32$$

Rewrite z^{2}+34z+64 as \left(z^{2}+2z\right)+\left(32z+64\right).

$$\left(z^{2}+2z\right)+\left(32z+64\right)$$

Factor out z in the first and 32 in the second group.

$$z\left(z+2\right)+32\left(z+2\right)$$

Factor out common term z+2 by using distributive property.

$$\left(z+2\right)\left(z+32\right)$$

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.

$$z^{2}+34z+64=0$$

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.

$$z=\frac{-34±\sqrt{34^{2}-4\times 64}}{2}$$

Square 34.

$$z=\frac{-34±\sqrt{1156-4\times 64}}{2}$$

Multiply -4 times 64.

$$z=\frac{-34±\sqrt{1156-256}}{2}$$

Add 1156 to -256.

$$z=\frac{-34±\sqrt{900}}{2}$$

Take the square root of 900.

$$z=\frac{-34±30}{2}$$

Now solve the equation z=\frac{-34±30}{2} when ± is plus. Add -34 to 30.

$$z=-\frac{4}{2}$$

Divide -4 by 2.

$$z=-2$$

Now solve the equation z=\frac{-34±30}{2} when ± is minus. Subtract 30 from -34.

$$z=-\frac{64}{2}$$

Divide -64 by 2.

$$z=-32$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2 for x_{1} and -32 for x_{2}.

$$z^{2}+34z+64=\left(z-\left(-2\right)\right)\left(z-\left(-32\right)\right)$$

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

$$z^{2}+34z+64=\left(z+2\right)\left(z+32\right)$$

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.

$$x ^ 2 +34x +64 = 0$$

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

$$r + s = -34 rs = 64$$

Two numbers r and s sum up to -34 exactly when the average of the two numbers is \frac{1}{2}*-34 = -17. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -17 - u s = -17 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 64

$$(-17 - u) (-17 + u) = 64$$

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

$$289 - u^2 = 64$$

Simplify the expression by subtracting 289 on both sides

$$-u^2 = 64-289 = -225$$

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

$$u^2 = 225 u = \pm\sqrt{225} = \pm 15$$

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

$$r =-17 - 15 = -32 s = -17 + 15 = -2$$

Evaluate

$$\left(z+2\right)\left(z+32\right)$$