Solve =4k^2+7k+3

$$= 4 k ^ { 2 } + 7 k + 3$$

Factor

$$\left(k+1\right)\left(4k+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4k^{2}+ak+bk+3. To find a and b, set up a system to be solved.

$$a+b=7 ab=4\times 3=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite 4k^{2}+7k+3 as \left(4k^{2}+3k\right)+\left(4k+3\right).

$$\left(4k^{2}+3k\right)+\left(4k+3\right)$$

Factor out k in 4k^{2}+3k.

$$k\left(4k+3\right)+4k+3$$

Factor out common term 4k+3 by using distributive property.

$$\left(4k+3\right)\left(k+1\right)$$

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.

$$4k^{2}+7k+3=0$$

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.

$$k=\frac{-7±\sqrt{7^{2}-4\times 4\times 3}}{2\times 4}$$

Square 7.

$$k=\frac{-7±\sqrt{49-4\times 4\times 3}}{2\times 4}$$

Multiply -4 times 4.

$$k=\frac{-7±\sqrt{49-16\times 3}}{2\times 4}$$

Multiply -16 times 3.

$$k=\frac{-7±\sqrt{49-48}}{2\times 4}$$

Add 49 to -48.

$$k=\frac{-7±\sqrt{1}}{2\times 4}$$

Take the square root of 1.

$$k=\frac{-7±1}{2\times 4}$$

Multiply 2 times 4.

$$k=\frac{-7±1}{8}$$

Now solve the equation k=\frac{-7±1}{8} when ± is plus. Add -7 to 1.

$$k=-\frac{6}{8}$$

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

$$k=-\frac{3}{4}$$

Now solve the equation k=\frac{-7±1}{8} when ± is minus. Subtract 1 from -7.

$$k=-\frac{8}{8}$$

Divide -8 by 8.

$$k=-1$$

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

$$4k^{2}+7k+3=4\left(k-\left(-\frac{3}{4}\right)\right)\left(k-\left(-1\right)\right)$$

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

$$4k^{2}+7k+3=4\left(k+\frac{3}{4}\right)\left(k+1\right)$$

Add \frac{3}{4} to k by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4k^{2}+7k+3=4\times \frac{4k+3}{4}\left(k+1\right)$$

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

$$4k^{2}+7k+3=\left(4k+3\right)\left(k+1\right)$$

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

$$x ^ 2 +\frac{7}{4}x +\frac{3}{4} = 0$$

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

$$r + s = -\frac{7}{4} rs = \frac{3}{4}$$

Two numbers r and s sum up to -\frac{7}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{4} = -\frac{7}{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>

$$r = -\frac{7}{8} - u s = -\frac{7}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{4}

$$(-\frac{7}{8} - u) (-\frac{7}{8} + u) = \frac{3}{4}$$

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

$$\frac{49}{64} - u^2 = \frac{3}{4}$$

Simplify the expression by subtracting \frac{49}{64} on both sides

$$-u^2 = \frac{3}{4}-\frac{49}{64} = -\frac{1}{64}$$

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

$$u^2 = \frac{1}{64} u = \pm\sqrt{\frac{1}{64}} = \pm \frac{1}{8}$$

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

$$r =-\frac{7}{8} - \frac{1}{8} = -1 s = -\frac{7}{8} + \frac{1}{8} = -0.750$$

Evaluate

$$\left(k+1\right)\left(4k+3\right)$$