# 1.1: Algebraic Simplification

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When algebraic techniques are presented as skills in isolation, they are much simpler to understand and practice. However the problem solving process in any context involves deciding which skills to use when. Most College Algebra students will have practiced problems in the form:
$\begin{array}{c} (x+7)(x-2)=? \\ \text { or } \\ (2 x+1)^{2}=? \end{array}$
The problems in this section deal with a combination of these processes which are often encountered as parts of more complex problems.

Example $$\PageIndex{1}$$

Simplify:
$3(x-1)(2 x+5)-(x+4)^{2} \nonumber$

Solution

In this example, the simplification involves two expressions: $$3(x-1)(2 x+5)$$ and $$(x+4)^{2} .$$ The $$(x+4)^{2}$$ is preceded by a negative (or subtraction) sign. This textbook will often treat $$-x$$ and $$+(-x)$$ as equivalent statements, since subtraction is defined as the addition of a negative.

We will simplify each expression separately and then look to combine like terms.
$3(x-1)(2 x+5)-(x+4)^{2}=3\left(2 x^{2}+3 x-5\right)-\left(x^{2}+8 x+16\right) \nonumber$
Notice that the results of both multiplications remain inside of parentheses. This is because each one has something that must be distributed.

In the case of $$\left(2 x^{2}+3 x-5\right)$$, there is a 3 which must be distributed, resulting in $$6 x^{2}+9 x-15 .$$ In the case of $$\left(x^{2}+8 x+16\right)$$ there is a negative sign or -1 which must be distributed, resulting in $$-x^{2}-8 x-16 .$$ It is important in these situations that the negative sign be distributed to all terms in the parentheses.

So
\begin{aligned} 3(x-1)(2 x+5)-(x+4)^{2} &=3\left(2 x^{2}+3 x-5\right)-\left(x^{2}+8 x+16\right) \\ &=6 x^{2}+9 x-15-x^{2}-8 x-16 \\ &=5 x^{2}+x-31 \end{aligned}

Example $$\PageIndex{2}$$

Simplify:
$2(x+3)^{2}-4(3 x-1)(x+2) \nonumber$

Solution

This example shows some of the same processes as the previous example. There are again two expressions that must be simplified, each of which has a coefficient that must be distributed. It is often helpful to wait until after multiplying the binomials before distributing the coefficient. However, as is often true in mathematics, there are several different approaches that may be taken in simplifying this problem.

If someone prefers to first distribute the coefficient before multiplying the binomials, then the coefficient must only be distributed to ONE of the binomials, but not both. For example, in multiplying $$3 * 2 * 5=30$$, we can first multiply $$2 * 5=10$$ and then $$3 * 10=30 .$$ Each factor is multiplied only once.
In the example above we can proceed as we did with the previous example:
\begin{aligned} 2(x+3)^{2}-4(3 x-1)(x+2) &=2\left(x^{2}+6 x+9\right)-4\left(3 x^{2}+5 x-2\right) \\ &=2 x^{2}+12 x+18-12 x^{2}-20 x+8 \\ &=-10 x^{2}-8 x+26 \end{aligned}
Or, we can choose to distribute the 4 first:
\begin{aligned} 2(x+3)^{2}-4(3 x-1)(x+2) &=2\left(x^{2}+6 x+9\right)-(12 x-4)(x+2) \\ &=2 x^{2}+12 x+18-\left(12 x^{2}+20 x-8\right) \\ &=2 x^{2}+12 x+18-12 x^{2}-20 x+8 \\ &=-10 x^{2}-8 x+26 \end{aligned}

Or, we can distribute the 4 as a negative. If we do this, then the sign in front of the parentheses will be positive:
\begin{aligned} 2(x+3)^{2}-4(3 x-1)(x+2) &=2\left(x^{2}+6 x+9\right)+(-12 x+4)(x+2) \\ &=2 x^{2}+12 x+18+\left(-12 x^{2}-20 x+8\right) \\ &=-10 x^{2}-8 x+26 \end{aligned}
Distributing the 2 in front of the squared binomial must also be handled carefully if you choose to do this. If you distribute the 2 before squaring the $$(x+3)$$, then the 2 will be squared as well. If you choose to distribute the $$2,$$ the $$(x+3)^{2}$$ must be written out as $$(x+3)(x+3)$$

\begin{aligned} 2(x+3)^{2}-4(3 x-1)(x+2) &=2(x+3)(x+3)-4(3 x-1)(x+2) \\ &=(2 x+6)(x+3)-4\left(3 x^{2}+5 x-2\right) \\ &=2 x^{2}+12 x+18-12 x^{2}-20 x+8 \\ &=-10 x^{2}-8 x+26 \end{aligned}

Most examples in this text will distribute the coefficients as the last step before combining like terms for a final answer.

Example $$\PageIndex{3}$$

Simplify:
$3 x[5-(2 x+7)]+(3 x-2)^{2}-(x-5)(x+4) \nonumber$

This example has three expressions that should be simplified separately before combining like terms. In the first expression $$3 x[5-(2 x+7)],$$ we should simplify inside the brackets before distributing the $$3 x$$

\begin{aligned}
& 3 x[5-(2 x+7)]+(3 x-2)^{2}-(x-5)(x+4) \\
=& 3 x[5-2 x-7]+(3 x-2)^{2}-(x-5)(x+4) \\
=& 3 x[-2 x-2]+(3 x-2)(3 x-2)-\left(x^{2}-x-20\right) \\
=&-6 x^{2}-6 x+\left(9 x^{2}-12 x+4\right)-x^{2}+x+20 \\
=&-6 x^{2}-6 x+9 x^{2}-12 x+4-x^{2}+x+20 \\
=& 2 x^{2}-17 x+24
\end{aligned}

Exercises $$\PageIndex{1}$$

Simplify each expression.

1. $$\quad (x-2)[2 x-2(3+x)]-(x+5)^{2}$$
2. $$\quad 3 x^{2}-[7 x-2(2 x-1)(3-x)]$$
3. $$\quad (a+b)^{2}-(a+b)(a-b)-\left[a(2 b-2)-\left(b^{2}-2 a\right)\right]$$
4. $$\quad 5 x-3(x-2)(x+7)+3(x-2)^{2}$$
5. $$\quad (m+3)(m-1)-(m-2)^{2}+4$$
6. $$\quad (a-1)(a-2)-(a-2)(a-3)+(a-3)(a-4)$$
7. $$\quad 2 a^{2}-3(a+1)(a-2)-[7-(a-1)]^{2}$$
8. $$\quad 2(x-5)(3 x+1)-(2 x-1)^{2}$$
9. $$\quad 6 y+(3 y+1)(y+2)-(y-3)(y-8)$$
10. $$\quad 6 x-4(x+10)(x-1)+(x+1)^{2}$$
1. $$\quad -x^{2}-16 x-13$$
2. $$\quad 3 b^{2}$$
3. $$\quad 6 m-3$$
4. $$\quad -2 a^{2}+19 a-58$$
5. $$\quad 2 y^{2}+24 y-22$$