Solve 3r^2-29r+56

$$3 r ^ { 2 } - 29 r + 56$$

Factor

$$\left(r-7\right)\left(3r-8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3r^{2}+ar+br+56. To find a and b, set up a system to be solved.

$$a+b=-29 ab=3\times 56=168$$

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 168.

$$-1,-168 -2,-84 -3,-56 -4,-42 -6,-28 -7,-24 -8,-21 -12,-14$$

Calculate the sum for each pair.

$$-1-168=-169 -2-84=-86 -3-56=-59 -4-42=-46 -6-28=-34 -7-24=-31 -8-21=-29 -12-14=-26$$

The solution is the pair that gives sum -29.

$$a=-21 b=-8$$

Rewrite 3r^{2}-29r+56 as \left(3r^{2}-21r\right)+\left(-8r+56\right).

$$\left(3r^{2}-21r\right)+\left(-8r+56\right)$$

Factor out 3r in the first and -8 in the second group.

$$3r\left(r-7\right)-8\left(r-7\right)$$

Factor out common term r-7 by using distributive property.

$$\left(r-7\right)\left(3r-8\right)$$

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.

$$3r^{2}-29r+56=0$$

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.

$$r=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 3\times 56}}{2\times 3}$$

Square -29.

$$r=\frac{-\left(-29\right)±\sqrt{841-4\times 3\times 56}}{2\times 3}$$

Multiply -4 times 3.

$$r=\frac{-\left(-29\right)±\sqrt{841-12\times 56}}{2\times 3}$$

Multiply -12 times 56.

$$r=\frac{-\left(-29\right)±\sqrt{841-672}}{2\times 3}$$

Add 841 to -672.

$$r=\frac{-\left(-29\right)±\sqrt{169}}{2\times 3}$$

Take the square root of 169.

$$r=\frac{-\left(-29\right)±13}{2\times 3}$$

The opposite of -29 is 29.

$$r=\frac{29±13}{2\times 3}$$

Multiply 2 times 3.

$$r=\frac{29±13}{6}$$

Now solve the equation r=\frac{29±13}{6} when ± is plus. Add 29 to 13.

$$r=\frac{42}{6}$$

Divide 42 by 6.

$$r=7$$

Now solve the equation r=\frac{29±13}{6} when ± is minus. Subtract 13 from 29.

$$r=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$r=\frac{8}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 7 for x_{1} and \frac{8}{3} for x_{2}.

$$3r^{2}-29r+56=3\left(r-7\right)\left(r-\frac{8}{3}\right)$$

Subtract \frac{8}{3} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3r^{2}-29r+56=3\left(r-7\right)\times \frac{3r-8}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3r^{2}-29r+56=\left(r-7\right)\left(3r-8\right)$$

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 3

$$x ^ 2 -\frac{29}{3}x +\frac{56}{3} = 0$$

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

$$r + s = \frac{29}{3} rs = \frac{56}{3}$$

Two numbers r and s sum up to \frac{29}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{29}{3} = \frac{29}{6}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{29}{6} - u s = \frac{29}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{56}{3}

$$(\frac{29}{6} - u) (\frac{29}{6} + u) = \frac{56}{3}$$

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

$$\frac{841}{36} - u^2 = \frac{56}{3}$$

Simplify the expression by subtracting \frac{841}{36} on both sides

$$-u^2 = \frac{56}{3}-\frac{841}{36} = -\frac{169}{36}$$

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

$$u^2 = \frac{169}{36} u = \pm\sqrt{\frac{169}{36}} = \pm \frac{13}{6}$$

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

$$r =\frac{29}{6} - \frac{13}{6} = 2.667 s = \frac{29}{6} + \frac{13}{6} = 7.000$$

Evaluate

$$\left(r-7\right)\left(3r-8\right)$$