Solve (4a^2-3a-7)=

$$( 4 a ^ { 2 } - 3 a - 7 ) =$$

Factor

$$\left(4a-7\right)\left(a+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4a^{2}+pa+qa-7. To find p and q, set up a system to be solved.

$$p+q=-3 pq=4\left(-7\right)=-28$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -28.

$$1,-28 2,-14 4,-7$$

Calculate the sum for each pair.

$$1-28=-27 2-14=-12 4-7=-3$$

The solution is the pair that gives sum -3.

$$p=-7 q=4$$

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

$$\left(4a^{2}-7a\right)+\left(4a-7\right)$$

Factor out a in 4a^{2}-7a.

$$a\left(4a-7\right)+4a-7$$

Factor out common term 4a-7 by using distributive property.

$$\left(4a-7\right)\left(a+1\right)$$

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.

$$4a^{2}-3a-7=0$$

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.

$$a=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 4\left(-7\right)}}{2\times 4}$$

Square -3.

$$a=\frac{-\left(-3\right)±\sqrt{9-4\times 4\left(-7\right)}}{2\times 4}$$

Multiply -4 times 4.

$$a=\frac{-\left(-3\right)±\sqrt{9-16\left(-7\right)}}{2\times 4}$$

Multiply -16 times -7.

$$a=\frac{-\left(-3\right)±\sqrt{9+112}}{2\times 4}$$

Add 9 to 112.

$$a=\frac{-\left(-3\right)±\sqrt{121}}{2\times 4}$$

Take the square root of 121.

$$a=\frac{-\left(-3\right)±11}{2\times 4}$$

The opposite of -3 is 3.

$$a=\frac{3±11}{2\times 4}$$

Multiply 2 times 4.

$$a=\frac{3±11}{8}$$

Now solve the equation a=\frac{3±11}{8} when ± is plus. Add 3 to 11.

$$a=\frac{14}{8}$$

Reduce the fraction \frac{14}{8} to lowest terms by extracting and canceling out 2.

$$a=\frac{7}{4}$$

Now solve the equation a=\frac{3±11}{8} when ± is minus. Subtract 11 from 3.

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

Divide -8 by 8.

$$a=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{4} for x_{1} and -1 for x_{2}.

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

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

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

Subtract \frac{7}{4} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Evaluate

$$\left(4a-7\right)\left(a+1\right)$$