Math Examples of Pre-Algebra

The zero of polynomial equation can be derived by equating P (y) to zero. ‘A’ is zero or root of polynomial P(X) if and only if P(X) = 0. For example,
Let P(X) = 5X³ − 4X² + 7X − 8.  Then one of the roots or zeros of this quadratic polynomial is 1. It can be proved by substituting the value y = 1 in the given polynomial.
P (1) = 5 (1)³  - 4 (1)² + 7 (1) – 8 = 5 – 4 + 7 – 8 = 0.
If (p + q i) is a root of the polynomial equation then (p – q i) is also a root of quadratic polynomial function.

Let’s consider an example in which if (2 + m) is a root to f (m) = - m 2 + 4 m – 5, then (2 – m) is also a root of the function.
Let’s put (2 + m) in the given equation
f (m) = - (2 + m) ² + 4 ( 2 + m) – 5,
        = - (4 – 1 + 4 m) + 3 + 4 m,
     = - 3 – 4 m + 3 + 4 m = 0,
f (m) = 0. If we substitute (2 – m) into the function, similar results will be obtained.

Example: If 5 is one zero of the quadratic polynomial x^2 minus kx minus 15 then value of k is?
Given polynomial equation is x² – k x – 15. This equation can be written in the form of a function as:
f (x) = x² – kx – 15,
In this problem, it is given to us that one zero exist at value of 'x' equals to 5. To prove this, substitute x = 5. This can be derived as:
f (5) = 5² – 5k – 15 = 0,
25 – 5k – 15 = 0,
10 – 5 k = 0,
5 k = 10 and hence k = 2.
Hence, if 5 is one zero of quadratic polynomial x² minus kx minus 15 then value of k is 2.

The above given polynomial can be written as a² – a – b² + b, = (a² – b²) – (a + b), = (a + b) * (a – b) – (a + b), Taking (a + b) common, we get: (a + b) * (a – b – 1).

The above polynomial will be  solved by the method of splitting.  First we will take 5 as the common factor of the three terms, so the polynomial will be  written as follows: = 5 *  (x ² + 6 * x + 9), = 5 *  (x ²  + 6 * x + 3²), = 5 *  (x ²  +  2 . 3 * x + 3²), Now we observe that the above expression is similar to the identity: (x + y)² = (x² + y² + 2 * x * y). So it can be written as: = 5 * (x + 3)².

Here we have 3 as common factor, so we will take out 3 common from both terms of the polynomial and we get: = 3 * (x² – 9) = 3 * (x² – 3²), Now we know the formula for a² – b² = (a + b) * (a – b), so we will apply the above identity  and we get : = 3 * (x – 3) * (x + 3).

We know  the identity (x² – y²) = (x + y) * (x – y),

Now here we will  look at the polynomial y⁴ – 81 and write it as :

Y⁴ – 3⁴,

= (y² – 3²) * (y² + 3²),

Again looking at (y² – 3²), it can be written as (y – 3) * (y + 3),

So we say that the expression y⁴ – 81 will be factorized into :

(y – 3) * (y + 3) * (y² + 3²).

Here we will first find prime factors of two Numbers 32 and 8. Thus we will get:
32 = 2 * 2 * 2 * 2 * 2,
8 = 2 * 2 * 2,
So we get :
√32 * √ 8 = √ (2 * 2 * 2 * 2 * 2) * √ (2 * 2 * 2),
√32 * √ 8 = (√ 2)⁸,
Or √32 * √ 8 = 2⁴,
Or √32 * √ 8 = 16.

In expression: (-1/4)^-3 * (-1/4)^-2, we observe that base of two Numbers is same, i.e. – ¼. Also the expression has two numbers with same base and different powers, with Operation of Multiplication. So we say that powers of the two will be added. Thus the above given expression can be written as: (-1/4)^-3-2,
= (-1/4)^-5,
= -4⁵,
= -1024.

In order to find the Square root of 625, first we will write the prime factors of the given number.
So we write: 625 = 5 * 5 * 5 * 5,
Now to find Square root of (625), we will make the pairs of the prime Numbers and we will pick one number from the Set.
On multiplying the resultant numbers, we will get the square root of the given number. So we get:
Square root (625) = 5 * 5 = 25.

The above given polynomial can be written as: (4 * x)² + (3y)² – 2 * 3 * 4 * x * y, Now we observe that the above given polynomial resembles the identity  (a – b⁾² = a² + b² – 2 * a * b, So we  write the factors of the polynomial = (4x – 3y)². 

Let us first assume that the required number will be 'x'. Thus we say that if the number 'x' is to be multiplied by the number (1 / 2)^-1, then we get the result as (-5/4)^-1,
This statement can be mathematically written as:
x * (1 / 2 )^-1 = (- 5 / 4)^-1,
In case any number is raised to the power of -1, then if we reverse the number, then the power becomes positive. Thus we come to the conclusion that the given expression can be written as :
x * (2 / 1) = (-4 / 5),
Or we can write : 2 * x = -4 / 5,
Or x = -4 / (5 * 2),
Or x = -20 / 2 = -10.

We know that  any number with the power 0 and any base, will result in number 1, so here we have 2⁰ = 1, 5⁰ = 1 and 4⁰ = 1,
Thus the expression (2⁰ * 5⁰) divided by (4⁰) becomes:
= (1 * 1) / 1,
= 1 /1 = 1.
Also we know that if 1 is multiplied by 1 gives 1 and the number 1 divided by 1 also gives 1. Thus we get:
= 1.