Solve 4x^2+24x+35

$$4 x ^ { 2 } + 24 x + 35$$

Factor

$$\left(2x+5\right)\left(2x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=24 ab=4\times 35=140$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 140.

$$1,140 2,70 4,35 5,28 7,20 10,14$$

Calculate the sum for each pair.

$$1+140=141 2+70=72 4+35=39 5+28=33 7+20=27 10+14=24$$

The solution is the pair that gives sum 24.

$$a=10 b=14$$

Rewrite 4x^{2}+24x+35 as \left(4x^{2}+10x\right)+\left(14x+35\right).

$$\left(4x^{2}+10x\right)+\left(14x+35\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(2x+5\right)+7\left(2x+5\right)$$

Factor out common term 2x+5 by using distributive property.

$$\left(2x+5\right)\left(2x+7\right)$$

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.

$$4x^{2}+24x+35=0$$

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.

$$x=\frac{-24±\sqrt{24^{2}-4\times 4\times 35}}{2\times 4}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\times 4\times 35}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-24±\sqrt{576-16\times 35}}{2\times 4}$$

Multiply -16 times 35.

$$x=\frac{-24±\sqrt{576-560}}{2\times 4}$$

Add 576 to -560.

$$x=\frac{-24±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$x=\frac{-24±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-24±4}{8}$$

Now solve the equation x=\frac{-24±4}{8} when ± is plus. Add -24 to 4.

$$x=-\frac{20}{8}$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{5}{2}$$

Now solve the equation x=\frac{-24±4}{8} when ± is minus. Subtract 4 from -24.

$$x=-\frac{28}{8}$$

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

$$x=-\frac{7}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{2} for x_{1} and -\frac{7}{2} for x_{2}.

$$4x^{2}+24x+35=4\left(x-\left(-\frac{5}{2}\right)\right)\left(x-\left(-\frac{7}{2}\right)\right)$$

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

$$4x^{2}+24x+35=4\left(x+\frac{5}{2}\right)\left(x+\frac{7}{2}\right)$$

Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+24x+35=4\times \frac{2x+5}{2}\left(x+\frac{7}{2}\right)$$

Add \frac{7}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+24x+35=4\times \frac{2x+5}{2}\times \frac{2x+7}{2}$$

Multiply \frac{2x+5}{2} times \frac{2x+7}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+24x+35=4\times \frac{\left(2x+5\right)\left(2x+7\right)}{2\times 2}$$

Multiply 2 times 2.

$$4x^{2}+24x+35=4\times \frac{\left(2x+5\right)\left(2x+7\right)}{4}$$

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

$$4x^{2}+24x+35=\left(2x+5\right)\left(2x+7\right)$$

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

$$x ^ 2 +6x +\frac{35}{4} = 0$$

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

$$r + s = -6 rs = \frac{35}{4}$$

Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. 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>

$$r = -3 - u s = -3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{35}{4}

$$(-3 - u) (-3 + u) = \frac{35}{4}$$

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

$$9 - u^2 = \frac{35}{4}$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = \frac{35}{4}-9 = -\frac{1}{4}$$

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.

$$r =-3 - \frac{1}{2} = -3.500 s = -3 + \frac{1}{2} = -2.500$$

Evaluate

$$\left(2x+5\right)\left(2x+7\right)$$