Solve 30j^2-27j-21

$$30 { j }^{ 2 } -27j-21$$

Factor

$$3\left(5j-7\right)\left(2j+1\right)$$

Solution Steps

Factor out 3.

$$3\left(10j^{2}-9j-7\right)$$

Consider 10j^{2}-9j-7. Factor the expression by grouping. First, the expression needs to be rewritten as 10j^{2}+aj+bj-7. To find a and b, set up a system to be solved.

$$a+b=-9 ab=10\left(-7\right)=-70$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -70.

$$1,-70 2,-35 5,-14 7,-10$$

Calculate the sum for each pair.

$$1-70=-69 2-35=-33 5-14=-9 7-10=-3$$

The solution is the pair that gives sum -9.

$$a=-14 b=5$$

Rewrite 10j^{2}-9j-7 as \left(10j^{2}-14j\right)+\left(5j-7\right).

$$\left(10j^{2}-14j\right)+\left(5j-7\right)$$

Factor out 2j in 10j^{2}-14j.

$$2j\left(5j-7\right)+5j-7$$

Factor out common term 5j-7 by using distributive property.

$$\left(5j-7\right)\left(2j+1\right)$$

Rewrite the complete factored expression.

$$3\left(5j-7\right)\left(2j+1\right)$$

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.

$$30j^{2}-27j-21=0$$

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.

$$j=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 30\left(-21\right)}}{2\times 30}$$

Square -27.

$$j=\frac{-\left(-27\right)±\sqrt{729-4\times 30\left(-21\right)}}{2\times 30}$$

Multiply -4 times 30.

$$j=\frac{-\left(-27\right)±\sqrt{729-120\left(-21\right)}}{2\times 30}$$

Multiply -120 times -21.

$$j=\frac{-\left(-27\right)±\sqrt{729+2520}}{2\times 30}$$

Add 729 to 2520.

$$j=\frac{-\left(-27\right)±\sqrt{3249}}{2\times 30}$$

Take the square root of 3249.

$$j=\frac{-\left(-27\right)±57}{2\times 30}$$

The opposite of -27 is 27.

$$j=\frac{27±57}{2\times 30}$$

Multiply 2 times 30.

$$j=\frac{27±57}{60}$$

Now solve the equation j=\frac{27±57}{60} when ± is plus. Add 27 to 57.

$$j=\frac{84}{60}$$

Reduce the fraction \frac{84}{60} to lowest terms by extracting and canceling out 12.

$$j=\frac{7}{5}$$

Now solve the equation j=\frac{27±57}{60} when ± is minus. Subtract 57 from 27.

$$j=-\frac{30}{60}$$

Reduce the fraction \frac{-30}{60} to lowest terms by extracting and canceling out 30.

$$j=-\frac{1}{2}$$

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

$$30j^{2}-27j-21=30\left(j-\frac{7}{5}\right)\left(j-\left(-\frac{1}{2}\right)\right)$$

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

$$30j^{2}-27j-21=30\left(j-\frac{7}{5}\right)\left(j+\frac{1}{2}\right)$$

Subtract \frac{7}{5} from j by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$30j^{2}-27j-21=30\times \frac{5j-7}{5}\left(j+\frac{1}{2}\right)$$

Add \frac{1}{2} to j by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$30j^{2}-27j-21=30\times \frac{5j-7}{5}\times \frac{2j+1}{2}$$

Multiply \frac{5j-7}{5} times \frac{2j+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$30j^{2}-27j-21=30\times \frac{\left(5j-7\right)\left(2j+1\right)}{5\times 2}$$

Multiply 5 times 2.

$$30j^{2}-27j-21=30\times \frac{\left(5j-7\right)\left(2j+1\right)}{10}$$

Cancel out 10, the greatest common factor in 30 and 10.

$$30j^{2}-27j-21=3\left(5j-7\right)\left(2j+1\right)$$

Evaluate

$$30j^{2}-27j-21$$