Solve 5t^2+15t+10

$$5 t ^ { 2 } + 15 t + 10$$

Factor

$$5\left(t+1\right)\left(t+2\right)$$

Solution Steps

Factor out 5.

$$5\left(t^{2}+3t+2\right)$$

Consider t^{2}+3t+2. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+2. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite t^{2}+3t+2 as \left(t^{2}+t\right)+\left(2t+2\right).

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

Factor out t in the first and 2 in the second group.

$$t\left(t+1\right)+2\left(t+1\right)$$

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

$$\left(t+1\right)\left(t+2\right)$$

Rewrite the complete factored expression.

$$5\left(t+1\right)\left(t+2\right)$$

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.

$$5t^{2}+15t+10=0$$

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.

$$t=\frac{-15±\sqrt{15^{2}-4\times 5\times 10}}{2\times 5}$$

Square 15.

$$t=\frac{-15±\sqrt{225-4\times 5\times 10}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-15±\sqrt{225-20\times 10}}{2\times 5}$$

Multiply -20 times 10.

$$t=\frac{-15±\sqrt{225-200}}{2\times 5}$$

Add 225 to -200.

$$t=\frac{-15±\sqrt{25}}{2\times 5}$$

Take the square root of 25.

$$t=\frac{-15±5}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{-15±5}{10}$$

Now solve the equation t=\frac{-15±5}{10} when ± is plus. Add -15 to 5.

$$t=-\frac{10}{10}$$

Divide -10 by 10.

$$t=-1$$

Now solve the equation t=\frac{-15±5}{10} when ± is minus. Subtract 5 from -15.

$$t=-\frac{20}{10}$$

Divide -20 by 10.

$$t=-2$$

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 -2 for x_{2}.

$$5t^{2}+15t+10=5\left(t-\left(-1\right)\right)\left(t-\left(-2\right)\right)$$

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

$$5t^{2}+15t+10=5\left(t+1\right)\left(t+2\right)$$

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 5

$$x ^ 2 +3x +2 = 0$$

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

$$r + s = -3 rs = 2$$

Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{2}. 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>

$$r = -\frac{3}{2} - u s = -\frac{3}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 2

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = 2$$

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

$$\frac{9}{4} - u^2 = 2$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = 2-\frac{9}{4} = -\frac{1}{4}$$

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

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

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

$$r =-\frac{3}{2} - \frac{1}{2} = -2 s = -\frac{3}{2} + \frac{1}{2} = -1$$

Evaluate

$$5\left(t+1\right)\left(t+2\right)$$