Good day my students, today you will learn how to operate perfect squares via factorization of algebraic expression.This mathematics involve expanding algebraic terms through factorizing variables.
We all know that (a + b)² is square of a and b,it is said to the square of sum of two quantities.This simply means,the variables in bracket is equal to the sum of their squares plus their products.This is waht i really mean:
(a + b)² = (a + b) (a + b) = a (a + b) + b (a + b)
= a² + ab + ab + b²
= a² +2ab + b²
therefore (a + b)² = a² +2ab + b²
Always remember that ab is the sum while b² is the product.
This operation involved two operators, plus ( + ) and minus ( - ) sign.
( a - b )² = ( a - b )( a - b ) = a( a - b ) - b( a - b )
= a² - ab - ab + b²
= a² -2ab + b²
therefore (a + b)² = a² -2ab + b²
In ( a - b )² , the square of difference of two quantities.This simply means,the variables in bracket is equal to the sum of their squares minus their products.
Example 1
Expand these expressions
a. (3m + 7n )² , b. (4u -5v)²
a. We say, (3m + 7n )² is equal to the ( square of 3m ) + (twice the product of 3m an 7n) +( the square of 7n). thus,
(3m + 7n )² = (3m)² + (2 x 3m x 7n) + (7n)²
= 9m² + 42mn + 49²
This is the same as (a² -2ab + b²) the first term and the last term with the power of 2.
Similarly (4u -5v)² = (4u)² + ( 2 x 4u x 5v ) + (5v)²
= 16u² - 40uv + 25²
To understand the above workings, you must know the basic operation/principles.
Write down the expressions of the following:
1. ( a + 4 )²
2.( 6 + f)²
3. ( 10h - 1)²
4. ( u + 5 )²
Solution
1. ( a + 4 )² = (a)² + ( 2 x a x 4 ) + (4)²
= a² + 8a + 16²
so therefore ( a + 4 )² = a² + 8a + 16²
2.( 6 + f)² = (6)² + ( 2 x f x 6 ) + (f)²
= 16² + 12f + f²
3. ( 10h - k)² = (10h)² + ( 2 x 1 x 10h) + (k)²
= 10h² - 20h + k²
4. ( u + 5 )² = (u)² + (2x u x 5 ) + (5)²
= u² + 10u + 25²
As you can see, the above operations are very simple,i think this is the most commonest algebraic mathematics so fare since the beginning of factorization.
Example 2
Now let discourse about factorization of quadratic expression.We all know that,a quadratic is the one whereby,2 is the highest power of the unknown variable(s) in the expression,Example, u² - 4x - 12, 63 - a² ,3y² + xy +10y² and so on.We say they are all quadratic expressions.
Since ( y + 2 )(y - 6 )
= y² -4y - 12
Break it down
This means, y x y = y²
+ 2y - 6 = - 4y
+ 2 x (-6) = -12
So therefore, ( y + 2 )(y - 6 ) = y² -4y - 12
( y + 2 ) and (y - 6 ) are the factors of (y² -4y - 12).In mathematics, 3 x 7 = 21,which means 3 and 7 are the factors of 21.Also,in real life two people can course problems and those people are called factors.
In arithmetic operations, 11 is said to be prime since it has no factor other than its self and 1.the same,
y² -4y - 12 has no factor (other than itself and 1).This rule affect all other prime numbers.
The secret behind this pattern is that the same number you add is the same you multiply.
Now factorize the following quadratic expressions.
1. d² + 6d + 5
2.a² + 14a + 13
3.c² + 8c + 15
4. y² + 10y + 16
Solution
1. d² + 6d + 5
= ( d + 1)(d +5)
Working
d x d = d²
(+1) + (+ 5d) = +6d ---+ x + = +
+1 x 5 = 5. ..The same number for addition and the same number for multiplication,i hope you understand my point.
2. a² + 14a + 13
= ( a + 13)( a + 13)
Working
a x a = a²
(+1) + (+ 13a) = 14a
+1 x 13 . The same number for addition and the same number for multiplication,i hope you understand my point.
3. c² + 8c + 15
= ( c + 3 )( c + 5)
Working
c x c = c²
(+3c) + ( +5) = +8
+3 x 5 = 15
The same number for addition and the same number for multiplication,i hope you understand my point.
That is how to solve problems on factorization of quadratic expression,please if you have questions put them in the comments box.
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