Expanding And Simplifying Algebraic Expressions A Comprehensive Guide

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In this article, we will delve into the process of expanding and simplifying algebraic expressions. We'll tackle expressions involving squares and products of polynomials, providing step-by-step solutions and explanations. Understanding these techniques is crucial for success in algebra and higher-level mathematics. We will address the following expressions:

  • (x−7)2(x-7)^2
  • (9s+7)2(9s + 7)^2
  • (t−4)(t2+t−6)(t-4)(t^2+t-6)
  • 4(r−5)(r+6)4(r-5)(r+6)

Understanding the Basics

Before we begin, let's quickly recap some essential algebraic concepts. Expanding an expression involves removing parentheses by applying the distributive property or using algebraic identities. Simplifying an expression means combining like terms and writing the expression in its most concise form. The goal is to present the expression in a way that is easier to understand and work with.

Key Concepts

  • Distributive Property: a(b+c)=ab+aca(b+c) = ab + ac
  • Squaring a Binomial: (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 and (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2
  • Combining Like Terms: Terms with the same variable and exponent can be added or subtracted.
  • Descending Order: Writing a polynomial with terms arranged from highest to lowest exponent.

Expanding (x−7)2(x-7)^2

To simplify the expression (x−7)2(x-7)^2, we will use the identity for the square of a binomial: (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2. In this case, a=xa = x and b=7b = 7. Applying the identity, we get:

(x−7)2=x2−2(x)(7)+72(x-7)^2 = x^2 - 2(x)(7) + 7^2

Now, let's simplify each term:

x2x^2 remains as x2x^2.

−2(x)(7)=−14x-2(x)(7) = -14x

72=497^2 = 49

Combining these terms, the simplified expression is:

x2−14x+49x^2 - 14x + 49

This quadratic expression is now in its simplest form, arranged in descending order of powers of xx.

Expanding (9s+7)2(9s + 7)^2

In this section, we will focus on expanding and simplifying the expression (9s+7)2(9s + 7)^2. This expression is the square of a binomial, similar to the previous example, but with slightly different terms. To tackle this, we'll again use the algebraic identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2. Here, a=9sa = 9s and b=7b = 7. Substituting these values into the identity gives us:

(9s+7)2=(9s)2+2(9s)(7)+72(9s + 7)^2 = (9s)^2 + 2(9s)(7) + 7^2

Now, let's break down each term and simplify:

  • (9s)2(9s)^2 means (9s)imes(9s)(9s) imes (9s). When we multiply these, we multiply the coefficients (9 and 9) and apply the exponent to the variable ss. Thus, (9s)2=81s2(9s)^2 = 81s^2.
  • The second term, 2(9s)(7)2(9s)(7), involves multiplying three factors together. We multiply the coefficients first: 2imes9imes7=1262 imes 9 imes 7 = 126. This gives us 126s126s, as ss is the variable term.
  • Finally, 727^2 is simply 7 multiplied by itself, which equals 49.

Putting these simplified terms together, we get:

81s2+126s+4981s^2 + 126s + 49

This resulting quadratic expression is now fully expanded and simplified. It's presented in descending order of the powers of ss, which is the standard form for quadratic expressions. The coefficient of s2s^2 is 81, the coefficient of ss is 126, and the constant term is 49. This form allows for easier analysis and further mathematical operations, such as factoring or solving equations.

Expanding (t−4)(t2+t−6)(t-4)(t^2+t-6)

Next, let's expand and simplify the expression (t−4)(t2+t−6)(t-4)(t^2+t-6). This involves multiplying a binomial (t−4)(t-4) by a trinomial (t2+t−6)(t^2+t-6). The key here is to apply the distributive property carefully. This means each term in the binomial must multiply each term in the trinomial. We'll take it step by step to ensure accuracy.

First, we multiply tt from the binomial by each term in the trinomial:

  • timest2=t3t imes t^2 = t^3 (when multiplying terms with the same base, we add the exponents)
  • timest=t2t imes t = t^2
  • times−6=−6tt imes -6 = -6t

So, multiplying tt by the trinomial gives us t3+t2−6tt^3 + t^2 - 6t.

Now, we multiply −4-4 from the binomial by each term in the trinomial:

  • −4imest2=−4t2-4 imes t^2 = -4t^2
  • −4imest=−4t-4 imes t = -4t
  • −4imes−6=24-4 imes -6 = 24 (a negative times a negative is positive)

Multiplying −4-4 by the trinomial yields −4t2−4t+24-4t^2 - 4t + 24.

Now, we combine these two results:

(t3+t2−6t)+(−4t2−4t+24)(t^3 + t^2 - 6t) + (-4t^2 - 4t + 24)

To simplify, we combine like terms. Like terms are terms that have the same variable raised to the same power:

  • We have one t3t^3 term.
  • For t2t^2 terms, we have t2t^2 and −4t2-4t^2. Combining these gives us 1t2−4t2=−3t21t^2 - 4t^2 = -3t^2.
  • For tt terms, we have −6t-6t and −4t-4t. Combining these gives us −6t−4t=−10t-6t - 4t = -10t.
  • The constant term is 24.

Putting it all together, the simplified expression is:

t3−3t2−10t+24t^3 - 3t^2 - 10t + 24

This polynomial is now in its simplest form, arranged in descending order of powers of tt. This process highlights the importance of careful distribution and combining like terms in algebraic simplification.

Expanding 4(r−5)(r+6)4(r-5)(r+6)

Finally, let's simplify the expression 4(r−5)(r+6)4(r-5)(r+6). This expression involves multiplying a constant (4) with the product of two binomials (r−5)(r-5) and (r+6)(r+6). The standard approach here is to first multiply the two binomials and then distribute the constant. This method helps in keeping the calculations organized and reduces the chances of errors.

First, we will multiply the binomials (r−5)(r-5) and (r+6)(r+6). This is done using the distributive property, often remembered by the acronym FOIL (First, Outer, Inner, Last), which guides us in multiplying each term of the first binomial with each term of the second binomial.

  • First: Multiply the first terms in each binomial: rimesr=r2r imes r = r^2.
  • Outer: Multiply the outer terms: rimes6=6rr imes 6 = 6r.
  • Inner: Multiply the inner terms: −5imesr=−5r-5 imes r = -5r.
  • Last: Multiply the last terms: −5imes6=−30-5 imes 6 = -30.

Combining these results gives us:

r2+6r−5r−30r^2 + 6r - 5r - 30

Now, we simplify by combining like terms. The like terms here are 6r6r and −5r-5r. Combining them gives us 6r−5r=r6r - 5r = r. So, the expression simplifies to:

r2+r−30r^2 + r - 30

This is the simplified product of the two binomials. Now, we need to multiply this trinomial by the constant 4. We distribute the 4 to each term in the trinomial:

  • 4imesr2=4r24 imes r^2 = 4r^2
  • 4imesr=4r4 imes r = 4r
  • 4imes−30=−1204 imes -30 = -120

So, the expression becomes:

4r2+4r−1204r^2 + 4r - 120

This is the final expanded and simplified form of the original expression 4(r−5)(r+6)4(r-5)(r+6). It's a quadratic expression presented in descending order of powers of rr, with the coefficients clearly displayed. This form is suitable for further analysis, such as finding roots or graphing the quadratic function.

Conclusion

In this article, we successfully expanded and simplified four algebraic expressions. We covered examples involving squaring binomials and multiplying polynomials. The key takeaways are the importance of the distributive property, algebraic identities, and combining like terms. Mastering these techniques is crucial for algebraic manipulation and problem-solving in mathematics. Remember to always double-check your work and ensure that the final expression is in its simplest form and arranged in descending order. With practice, these skills will become second nature, making algebra more accessible and enjoyable.