Solve {l}{4t^2+32t}{-36}

$$\left. \begin{array} { l } { 4 t ^ { 2 } + 32 t } \\ { - 36 } \end{array} \right.$$

Factor

$$4\left(t-1\right)\left(t+9\right)$$

Solution Steps

Factor out 4.

$$4\left(t^{2}+8t-9\right)$$

Consider t^{2}+8t-9. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-9. To find a and b, set up a system to be solved.

$$a+b=8 ab=1\left(-9\right)=-9$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -9.

$$-1,9 -3,3$$

Calculate the sum for each pair.

$$-1+9=8 -3+3=0$$

The solution is the pair that gives sum 8.

$$a=-1 b=9$$

Rewrite t^{2}+8t-9 as \left(t^{2}-t\right)+\left(9t-9\right).

$$\left(t^{2}-t\right)+\left(9t-9\right)$$

Factor out t in the first and 9 in the second group.

$$t\left(t-1\right)+9\left(t-1\right)$$

Factor out common term t-1 by using distributive property.

$$\left(t-1\right)\left(t+9\right)$$

Rewrite the complete factored expression.

$$4\left(t-1\right)\left(t+9\right)$$

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.

$$4t^{2}+32t-36=0$$

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.

$$t=\frac{-32±\sqrt{32^{2}-4\times 4\left(-36\right)}}{2\times 4}$$

Square 32.

$$t=\frac{-32±\sqrt{1024-4\times 4\left(-36\right)}}{2\times 4}$$

Multiply -4 times 4.

$$t=\frac{-32±\sqrt{1024-16\left(-36\right)}}{2\times 4}$$

Multiply -16 times -36.

$$t=\frac{-32±\sqrt{1024+576}}{2\times 4}$$

Add 1024 to 576.

$$t=\frac{-32±\sqrt{1600}}{2\times 4}$$

Take the square root of 1600.

$$t=\frac{-32±40}{2\times 4}$$

Multiply 2 times 4.

$$t=\frac{-32±40}{8}$$

Now solve the equation t=\frac{-32±40}{8} when ± is plus. Add -32 to 40.

$$t=\frac{8}{8}$$

Divide 8 by 8.

$$t=1$$

Now solve the equation t=\frac{-32±40}{8} when ± is minus. Subtract 40 from -32.

$$t=-\frac{72}{8}$$

Divide -72 by 8.

$$t=-9$$

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 -9 for x_{2}.

$$4t^{2}+32t-36=4\left(t-1\right)\left(t-\left(-9\right)\right)$$

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

$$4t^{2}+32t-36=4\left(t-1\right)\left(t+9\right)$$

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

$$x ^ 2 +8x -9 = 0$$

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

$$r + s = -8 rs = -9$$

Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -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>

$$r = -4 - u s = -4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -9

$$(-4 - u) (-4 + u) = -9$$

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

$$16 - u^2 = -9$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = -9-16 = -25$$

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

$$u^2 = 25 u = \pm\sqrt{25} = \pm 5$$

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

$$r =-4 - 5 = -9 s = -4 + 5 = 1$$

Evaluate

$$4\left(t-1\right)\left(t+9\right)$$