Algebra 1 Final Exam Practice

Introduction to algebra 1 final exam practice:

Let us see about algebra connections equations. The algebra is a mathematics that can be studied in words of mathematics numbers. In this we learn about the algebra connections equations with a few concepts. The algebra is the branch in which we learned the arithmetical functions, polynomials, factors and some equations. Let us see some examples in  algebra 1 final exam practice.

Important Concepts of Algebra

The followings are some of the algebra concepts.

1. Polynomial

In algebra connections, a polynomial of degree one is known as a linear polynomial.

2. Quadratic equation

In algebra connections, a quadratic equations in a variable of the x .An equation in the form ax2 + bx + c = 0.

3. Algebraic identities

The form (a + b)2 = a2 + 2ab + b2 is called as the algebraic identities.

4. Algebraic expressions

It is a division of polynomial can be denoted by a letters and exponent.


Examples to understand algebra 1 final exam practice:


Algebra Connections Problems

1)Solve the following equation 49x2 – 36 = 0.

Solution:

49x2 – 36 = 0 or

(7x + 6) (7x – 6) = 0 or (by simplify the equations)

7x2- 62 =0

7x2- 36 = 0

x2= 36/7

The solution set = `6 / sqrt 7`

2) Conclude whether (x–3) is a cause of polynomial p(x) = x³ – 3x² + 5x – 15.

Solution:

For (x–3) is a factor of p(x),

p(3) must be zero by the factor theorem.

Now p(3) = 3³ – 3(3) ² + 5(3) – 15

= 27 – 27 + 15 – 15 (by equating the known polynomial)

= 0

Hence (x–3) is a factor of the given polynomial.

3) Calculate 105 * 104 without multiplying directly.

Solution:

105 * 106 = (100 + 5) * (100 + 4)

= (100)² + (4 + 4) (100) + (5 * 4) (by using identity)

= 10000 + 800 + 20

= 10820.

4) Expand: (3a + 2b – 3c) ²

Solution:

Comparing the given expression (3a + 2b – 3c)² with (a + b + c) ²,

From the expression we have,

a = 3a

b = 2b

c = -3c

(3a + 2b – 3c) ² = (3a) ²+ (2b) ² - (3c) ²+ 2(3a) (2b) + 2(2b)(-3c) + 2(-3c)(3a)

Therefore, the equation becomes as,

= 9a² + 4b² + 9c² + 12ab² -12bc -18ca.

These are the examples of algebra connections equations.

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Exam questions to practice for algebra 1 final exam practice:

Problem 1:

Solve the equation 6x + 8 = 38 for x.

Answer: 5

Problem 2:

Solve the equation 8x + (40 ÷ 8) = 53 for x.

Answer: x = 6

Problem 3:

Solve the equation for x2 + 8= 89 for x.

Answer: x = 9

Problem 4:

Solve the equation 5x2 + 15x + 10 = 0 by using quadratic formula.

Answer: x1 = -1 and x2 = -2

Problem 5:

Solve the equation 5x2 + 15x + 10 = 0 by using factorizing method.

Answer: x1 = -1 and x2 = -2

Problem 6:

Solve the equation 3x2 + 4x + 1 = 0 by using quadratic formula.

Answer: x1 = -0.3 and x2 = -1

Problem 7:

Solve the equation 3x2 + 4x + 1 = 0 by using factorizing method.

Answer: x1 = -0.3 and x2 = -1

These are algebra 1 final exam practice as for exam preparation.

Factoring Polynomials by Grouping

Introduction of factoring polynomials by grouping:

• In general, the sum of a countable number of monomials is referred as a polynomials. The way of writing a polynomials as a product of two or more simpler polynomials is calledfactorization. The process of factorization is also known as the resolution into factors or factoring polynomials. Each simpler polynomial in the product is called a factor of the given polynomial.   
• Factoring polynomial by grouping method is used for the expressions containing more than three terms.

Factoring Polynomials Using Grouping:

Factoring polynomials by grouping involving four or more than four terms can be grouped using the following steps:

Step 1:  The terms having common factors should be grouped.

Step 2: The Greatest Common Factor (GCF) is taken out.

Step 3:  In the third step, we have to use the distributive law to find the factors.

                                               The distributive law is given by

                                                         a (b + c) = a b + a c

Example Problems on Factoring Polynomials by Grouping:

Ex 1:  Factoring the given polynomial, 3x² + 9x³ + 7x⁷ + 21x⁸, by grouping:     

Sol :   Step 1:  The terms having common factors should be grouped.

                                     (3x² + 9x³⁾ + (7x⁷ + 21x⁸).

Step 2: The Greatest Common Factor (GCF) is taken out.

                            3x² + 9x³ and 7x⁷ + 21x⁸ both can have a GCF.

               Taking out the GCF outside, we get

                         (3x² + 9x³⁾+ (7x⁷ + 21x⁸) = 3x² (1 + 3x) + 7x⁷ (1 + 3x)

Step 3: Use  distributive law to find the factors.

                Note that there is a common factor, 1 + 3x. So, factor out 1+3x.

                  By distributive law, we get

                                3x² + 9x³ + 7x⁷ + 21x⁸ = (3x² + 7x⁷) (1 + 3x).

Ex  2:   Factoring the given polynomial by grouping:  2x³+6x²−6x−18

Sol :   Step 1: The terms having common factors should be grouped.

                                               (2x³+6x²) + (−6x−18)

Step 2: The Greatest Common Factor (GCF) is taken out.

                                  (2x³+6x²) and  (−6x−18) both have a GCF

               Taking out the GCF outside, we get

                                   (2x³+6x²) + (−6x−18) = 2x²(x+3)-6(x+3)

Step 3:  Use distributive law to find the factors.

              Note that there is a common factor, x+3. So, Factor out (x+3).

               By distributive law, we get

                                             2x³+6x²−6x−18=(x +3) (2x²-6).

Ex 3 :  Factoring the given expression by grouping: 2x² - 3x + 10x - 15.

Sol :  Step 1:  The terms having common factors should be grouped.

                                       (2x² + 10x) + (- 3x – 15)

Step 2: The Greatest Common Factor (GCF) is taken out.

                            2x²+10x) and  (−3x−15) both have a GCF

              Taking out the GCF outside, we get

                              (2x²+10x) + (−3x−15) = 2x(x+5)-3(x+5)

Step 3:  Use distributive law to find the factors.

              Note that there is a common factor, x+5. So, Factor out (x+5).

               By distributive law,we get

                                             2x² - 3x + 10x – 15=(x+5) (2x-3).

These are the examples for the factoring polynomials by grouping.

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