Pre algebra simplify steps
Introduction :
Pre algebra is a common name for a course in middle school student mathematics. Pre algebra includes several broad subjects are Review of natural number arithmetic, new types of numbers such as integers, fractions, decimals and negative numbers, factorization of natural number, Rules of evaluation of expressions, such as operator precedence and use of parentheses . Basics of the equations, including rules for invariant manipulation of equations Variables and exponentiation. (Source.Wikipedia)
Examples for pre algebra simplify steps:
Additions of Polynomials simplify steps:
We this method add two polynomial by adding the coefficients of the like power.
Example 1:
Find the sum of 17x4 – 3x2 + 5x + 3 and 8x + 6x3 – 6x2 – 1.
Solution:
Using the associative and distributive properties of the real numbers, we obtain
(17x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 8x – 1) = 17x4 + 6x3 – 3x2 – 6x2 + 5x + 8x + 3 – 1
= 17x4 + 6x3 – (3+6)x2 + (5+8)x + 2
= 17x4 + 6x3 – 9x2 + 13x + 2.
Subtractions of Polynomials simplify steps:
We subtract polynomials like addition of polynomials.
Example 2:
Subtract 2x3 – 3x2 – 9 from 5x3 + 5x2 – 8x – 6.
Solution:
Using associative and distributive properties,
(5x3 + 5x2 – 8x – 6) – (2x3 – 3x2 – 9) = 5x3 + 5x2 – 8x – 6 – 2x3 + 3x2 + 9
= 5x3 – 2x3 + 5x2 + 3x2 – 8x – 6 + 9
= (5x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+9)
= 3x3 + 8x2 – 8x – 3.
Multiplications of two polynomials simplify steps:
To find the multiplication or product of the two polynomials, we use the distributive properties and the law of exponents.
Example 3:
Find the product of 3x3 – 2x2 – 6 and 2x2 + 3x – 1.
Solution: (3x3 – 2x2 – 6) (2x2 + 3x – 1)
= 3x3 (2x2 + 3x – 1) + (–2x2) (2x2 + 3x – 1) + (–6) (2x2 + 3x – 1)
= (6x5 + 9x4 – 3x3) + (–4x4 – 6x3 + 2x2) + (–12x2 – 18x + 6)
= 6x5 + 9x4 –3 x3 – 4x4 – 6x3 + 2x2 – 12x2 – 18x + 6
= 6x5 + (9x4 – 4x4) + (–3x3 – 6x3) + (2x2 – 12x2) + (–18x) +6
= 4x5 + 5x4 – 9x3 – 10x2 – 18x + 6.
Factorize the pre algebra by simple steps:
Example 1:
Factorize 49m2 - 625n2
Solution:
We have 492- 625n2 = (7m)2- (25n)2
= (7m+25n) (7m-25n)
Example 2:
Factorize 9a2-(x-y)2
Solution:
We have
= [3a+(x-y)] [3a-(x-y)]
= (3a+x-y) (3a-x+y)
Practice problem for pre algebra simplify steps:
1. Find the sum and write it in the standard form:
(2x4 + x2 + 3x) + (x4 – 3x2 + 9x – 8)
Answer: 3x4 -2x2+12x-8
2. Find the subtraction and write it in the standard form:
(x4 – 3x2 + 7x – 80) – (3x4 + x2 + 3x)
Answer: -2x4-4x2+4x-80
Pre algebra is a common name for a course in middle school student mathematics. Pre algebra includes several broad subjects are Review of natural number arithmetic, new types of numbers such as integers, fractions, decimals and negative numbers, factorization of natural number, Rules of evaluation of expressions, such as operator precedence and use of parentheses . Basics of the equations, including rules for invariant manipulation of equations Variables and exponentiation. (Source.Wikipedia)
Examples for pre algebra simplify steps:
Additions of Polynomials simplify steps:
We this method add two polynomial by adding the coefficients of the like power.
Example 1:
Find the sum of 17x4 – 3x2 + 5x + 3 and 8x + 6x3 – 6x2 – 1.
Solution:
Using the associative and distributive properties of the real numbers, we obtain
(17x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 8x – 1) = 17x4 + 6x3 – 3x2 – 6x2 + 5x + 8x + 3 – 1
= 17x4 + 6x3 – (3+6)x2 + (5+8)x + 2
= 17x4 + 6x3 – 9x2 + 13x + 2.
Subtractions of Polynomials simplify steps:
We subtract polynomials like addition of polynomials.
Example 2:
Subtract 2x3 – 3x2 – 9 from 5x3 + 5x2 – 8x – 6.
Solution:
Using associative and distributive properties,
(5x3 + 5x2 – 8x – 6) – (2x3 – 3x2 – 9) = 5x3 + 5x2 – 8x – 6 – 2x3 + 3x2 + 9
= 5x3 – 2x3 + 5x2 + 3x2 – 8x – 6 + 9
= (5x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+9)
= 3x3 + 8x2 – 8x – 3.
Multiplications of two polynomials simplify steps:
To find the multiplication or product of the two polynomials, we use the distributive properties and the law of exponents.
Example 3:
Find the product of 3x3 – 2x2 – 6 and 2x2 + 3x – 1.
Solution: (3x3 – 2x2 – 6) (2x2 + 3x – 1)
= 3x3 (2x2 + 3x – 1) + (–2x2) (2x2 + 3x – 1) + (–6) (2x2 + 3x – 1)
= (6x5 + 9x4 – 3x3) + (–4x4 – 6x3 + 2x2) + (–12x2 – 18x + 6)
= 6x5 + 9x4 –3 x3 – 4x4 – 6x3 + 2x2 – 12x2 – 18x + 6
= 6x5 + (9x4 – 4x4) + (–3x3 – 6x3) + (2x2 – 12x2) + (–18x) +6
= 4x5 + 5x4 – 9x3 – 10x2 – 18x + 6.
Factorize the pre algebra by simple steps:
Example 1:
Factorize 49m2 - 625n2
Solution:
We have 492- 625n2 = (7m)2- (25n)2
= (7m+25n) (7m-25n)
Example 2:
Factorize 9a2-(x-y)2
Solution:
We have
= [3a+(x-y)] [3a-(x-y)]
= (3a+x-y) (3a-x+y)
Practice problem for pre algebra simplify steps:
1. Find the sum and write it in the standard form:
(2x4 + x2 + 3x) + (x4 – 3x2 + 9x – 8)
Answer: 3x4 -2x2+12x-8
2. Find the subtraction and write it in the standard form:
(x4 – 3x2 + 7x – 80) – (3x4 + x2 + 3x)
Answer: -2x4-4x2+4x-80