Solve t^2-7t+10

$$t ^ { 2 } - 7 t + 10$$

Factor

$$\left(t-5\right)\left(t-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+10. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 10=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite t^{2}-7t+10 as \left(t^{2}-5t\right)+\left(-2t+10\right).

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

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

$$t\left(t-5\right)-2\left(t-5\right)$$

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

$$\left(t-5\right)\left(t-2\right)$$

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.

$$t^{2}-7t+10=0$$

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.

$$t=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 10}}{2}$$

Square -7.

$$t=\frac{-\left(-7\right)±\sqrt{49-4\times 10}}{2}$$

Multiply -4 times 10.

$$t=\frac{-\left(-7\right)±\sqrt{49-40}}{2}$$

Add 49 to -40.

$$t=\frac{-\left(-7\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$t=\frac{-\left(-7\right)±3}{2}$$

The opposite of -7 is 7.

$$t=\frac{7±3}{2}$$

Now solve the equation t=\frac{7±3}{2} when ± is plus. Add 7 to 3.

$$t=\frac{10}{2}$$

Divide 10 by 2.

$$t=5$$

Now solve the equation t=\frac{7±3}{2} when ± is minus. Subtract 3 from 7.

$$t=\frac{4}{2}$$

Divide 4 by 2.

$$t=2$$

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 2 for x_{2}.

$$t^{2}-7t+10=\left(t-5\right)\left(t-2\right)$$

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.

$$x ^ 2 -7x +10 = 0$$

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

$$r + s = 7 rs = 10$$

Two numbers r and s sum up to 7 exactly when the average of the two numbers is \frac{1}{2}*7 = \frac{7}{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>

$$r = \frac{7}{2} - u s = \frac{7}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 10

$$(\frac{7}{2} - u) (\frac{7}{2} + u) = 10$$

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

$$\frac{49}{4} - u^2 = 10$$

Simplify the expression by subtracting \frac{49}{4} on both sides

$$-u^2 = 10-\frac{49}{4} = -\frac{9}{4}$$

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

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

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

$$r =\frac{7}{2} - \frac{3}{2} = 2 s = \frac{7}{2} + \frac{3}{2} = 5$$

Evaluate

$$\left(t-5\right)\left(t-2\right)$$