Solve 4t^2-32t+63

$$4 t ^ { 2 } - 32 t + 63$$

Factor

$$\left(2t-9\right)\left(2t-7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4t^{2}+at+bt+63. To find a and b, set up a system to be solved.

$$a+b=-32 ab=4\times 63=252$$

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 252.

$$-1,-252 -2,-126 -3,-84 -4,-63 -6,-42 -7,-36 -9,-28 -12,-21 -14,-18$$

Calculate the sum for each pair.

$$-1-252=-253 -2-126=-128 -3-84=-87 -4-63=-67 -6-42=-48 -7-36=-43 -9-28=-37 -12-21=-33 -14-18=-32$$

The solution is the pair that gives sum -32.

$$a=-18 b=-14$$

Rewrite 4t^{2}-32t+63 as \left(4t^{2}-18t\right)+\left(-14t+63\right).

$$\left(4t^{2}-18t\right)+\left(-14t+63\right)$$

Factor out 2t in the first and -7 in the second group.

$$2t\left(2t-9\right)-7\left(2t-9\right)$$

Factor out common term 2t-9 by using distributive property.

$$\left(2t-9\right)\left(2t-7\right)$$

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.

$$4t^{2}-32t+63=0$$

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.

$$t=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 4\times 63}}{2\times 4}$$

Square -32.

$$t=\frac{-\left(-32\right)±\sqrt{1024-4\times 4\times 63}}{2\times 4}$$

Multiply -4 times 4.

$$t=\frac{-\left(-32\right)±\sqrt{1024-16\times 63}}{2\times 4}$$

Multiply -16 times 63.

$$t=\frac{-\left(-32\right)±\sqrt{1024-1008}}{2\times 4}$$

Add 1024 to -1008.

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

Take the square root of 16.

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

The opposite of -32 is 32.

$$t=\frac{32±4}{2\times 4}$$

Multiply 2 times 4.

$$t=\frac{32±4}{8}$$

Now solve the equation t=\frac{32±4}{8} when ± is plus. Add 32 to 4.

$$t=\frac{36}{8}$$

Reduce the fraction \frac{36}{8} to lowest terms by extracting and canceling out 4.

$$t=\frac{9}{2}$$

Now solve the equation t=\frac{32±4}{8} when ± is minus. Subtract 4 from 32.

$$t=\frac{28}{8}$$

Reduce the fraction \frac{28}{8} to lowest terms by extracting and canceling out 4.

$$t=\frac{7}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{2} for x_{1} and \frac{7}{2} for x_{2}.

$$4t^{2}-32t+63=4\left(t-\frac{9}{2}\right)\left(t-\frac{7}{2}\right)$$

Subtract \frac{9}{2} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4t^{2}-32t+63=4\times \frac{2t-9}{2}\left(t-\frac{7}{2}\right)$$

Subtract \frac{7}{2} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4t^{2}-32t+63=4\times \frac{2t-9}{2}\times \frac{2t-7}{2}$$

Multiply \frac{2t-9}{2} times \frac{2t-7}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4t^{2}-32t+63=4\times \frac{\left(2t-9\right)\left(2t-7\right)}{2\times 2}$$

Multiply 2 times 2.

$$4t^{2}-32t+63=4\times \frac{\left(2t-9\right)\left(2t-7\right)}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4t^{2}-32t+63=\left(2t-9\right)\left(2t-7\right)$$

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

$$x ^ 2 -8x +\frac{63}{4} = 0$$

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

$$r + s = 8 rs = \frac{63}{4}$$

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 = \frac{63}{4}

$$(4 - u) (4 + u) = \frac{63}{4}$$

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

$$16 - u^2 = \frac{63}{4}$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = \frac{63}{4}-16 = -\frac{1}{4}$$

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.

$$r =4 - \frac{1}{2} = 3.500 s = 4 + \frac{1}{2} = 4.500$$

Evaluate

$$\left(2t-9\right)\left(2t-7\right)$$