Solve 4t^2-43t+30

$$4 t ^ { 2 } - 43 t + 30$$

Factor

$$\left(t-10\right)\left(4t-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4t^{2}+at+bt+30. To find a and b, set up a system to be solved.

$$a+b=-43 ab=4\times 30=120$$

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 120.

$$-1,-120 -2,-60 -3,-40 -4,-30 -5,-24 -6,-20 -8,-15 -10,-12$$

Calculate the sum for each pair.

$$-1-120=-121 -2-60=-62 -3-40=-43 -4-30=-34 -5-24=-29 -6-20=-26 -8-15=-23 -10-12=-22$$

The solution is the pair that gives sum -43.

$$a=-40 b=-3$$

Rewrite 4t^{2}-43t+30 as \left(4t^{2}-40t\right)+\left(-3t+30\right).

$$\left(4t^{2}-40t\right)+\left(-3t+30\right)$$

Factor out 4t in the first and -3 in the second group.

$$4t\left(t-10\right)-3\left(t-10\right)$$

Factor out common term t-10 by using distributive property.

$$\left(t-10\right)\left(4t-3\right)$$

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.

$$4t^{2}-43t+30=0$$

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.

$$t=\frac{-\left(-43\right)±\sqrt{\left(-43\right)^{2}-4\times 4\times 30}}{2\times 4}$$

Square -43.

$$t=\frac{-\left(-43\right)±\sqrt{1849-4\times 4\times 30}}{2\times 4}$$

Multiply -4 times 4.

$$t=\frac{-\left(-43\right)±\sqrt{1849-16\times 30}}{2\times 4}$$

Multiply -16 times 30.

$$t=\frac{-\left(-43\right)±\sqrt{1849-480}}{2\times 4}$$

Add 1849 to -480.

$$t=\frac{-\left(-43\right)±\sqrt{1369}}{2\times 4}$$

Take the square root of 1369.

$$t=\frac{-\left(-43\right)±37}{2\times 4}$$

The opposite of -43 is 43.

$$t=\frac{43±37}{2\times 4}$$

Multiply 2 times 4.

$$t=\frac{43±37}{8}$$

Now solve the equation t=\frac{43±37}{8} when ± is plus. Add 43 to 37.

$$t=\frac{80}{8}$$

Divide 80 by 8.

$$t=10$$

Now solve the equation t=\frac{43±37}{8} when ± is minus. Subtract 37 from 43.

$$t=\frac{6}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$t=\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and \frac{3}{4} for x_{2}.

$$4t^{2}-43t+30=4\left(t-10\right)\left(t-\frac{3}{4}\right)$$

Subtract \frac{3}{4} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4t^{2}-43t+30=4\left(t-10\right)\times \frac{4t-3}{4}$$

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

$$4t^{2}-43t+30=\left(t-10\right)\left(4t-3\right)$$

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

$$x ^ 2 -\frac{43}{4}x +\frac{15}{2} = 0$$

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

$$r + s = \frac{43}{4} rs = \frac{15}{2}$$

Two numbers r and s sum up to \frac{43}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{43}{4} = \frac{43}{8}. 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>

$$r = \frac{43}{8} - u s = \frac{43}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{15}{2}

$$(\frac{43}{8} - u) (\frac{43}{8} + u) = \frac{15}{2}$$

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

$$\frac{1849}{64} - u^2 = \frac{15}{2}$$

Simplify the expression by subtracting \frac{1849}{64} on both sides

$$-u^2 = \frac{15}{2}-\frac{1849}{64} = -\frac{1369}{64}$$

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

$$u^2 = \frac{1369}{64} u = \pm\sqrt{\frac{1369}{64}} = \pm \frac{37}{8}$$

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

$$r =\frac{43}{8} - \frac{37}{8} = 0.750 s = \frac{43}{8} + \frac{37}{8} = 10$$

Evaluate

$$\left(t-10\right)\left(4t-3\right)$$