The document defines and provides examples of polynomials. It discusses that a polynomial is an algebraic expression with terms of increasing powers of a variable and constant coefficients. The degree of a polynomial is the highest power of its terms. Zeros or roots are numbers that make the polynomial equal to zero. The document provides methods for factorizing polynomials by splitting middle terms and finding all factors to determine all zeros. It also establishes relationships between the coefficients and zeros of a quadratic polynomial.

2. A polynomialp(x) in one variable x is an algebraic expression in x of
the form
p(x) = anxn + an–1xn – 1 + . . . + a2x2 + a1x + a0,
where a
0, a1, a2, . . ., an are constants and an ≠ 0.
a
0, a1, a2, . . ., an are respectively the coefficients of x0, x, x2, . . ., xn,
and n is called the degree
of the polynomial. Each ofanxn, an–1 xn–1, ..., a0, with an≠ 0, is
called a term of the polynomial
p(x).

3. The degree of a polynomial is the highest power
of it terms
ex- 2x2+6x+12 (degree = 2)
Polynomials of degrees
1- linear polynomial
2- quadratic polynomial
3- cubic polynomial

4. A zero or root of a polynomial is a number that,
when plugged in for the variable, makes the
function equal to zero . To fide the zero of the
polynomial just make factor of it by splitting the
middle term

5. Ex- 12x2 – 7x + 1
Here p+q = - 7 , pq = 12
= 12 × 1 = 12
Therefore p + q = – 7 = – 4 – 3
and pq = 12 = (– 4)(–3)
Therefore
12x2 – 7x + 1= 12x2+(– 4 – 3)x+1
= 12x2 – 4x – 3x +1
=4x(3x-1)-1(3x-1)
=(4x-1)(3x-1)

6. (4x-1)(3x-1)
Therefore
X= 1/4
X= 1/3
Therefore zeros are are 1/4 1/3
It means the polynomial f(x) is equals to
zero when x is 1/3 1/4

7. Ex= x3 + 13x2+ 32x + 20
here we hat to split two middle terms 13x2 ,32x
We can write 13x2 =x2 + 12x2
32x =12x + 20x
therefore x3 + 13x2+ 32x + 20= x3+x2+12x2 +12x+20x+20
=x2(x+1) 12x(x+1) 20(x+1)
=(x+1)(x2 +12x+20)

8. Now we have two factors (x+1)(x2 +12x+20) but f(x) is cubic
polynomial ,so it has three factor
Now we make factors of (x2 +12x+20)
(x2 +12x+20)= x2 +2x+10x+20
= x(x+2)10(x+2)
= (x+10)(x+2)
= (x+1) (x+10)(x+2) are factor of f(x)
Therefore
X=-1,-10,-2
It means the polynomial f(x) is equals to zero when x is 1,-10,-2

9. In general, ifα and β are the zeroes of the quadratic polynomialp(x) =
ax2+ bx + c,
a ≠ 0, then you know that x – α and x – β are the factors of p(x).
Therefore,
ax2 + bx + c = k(x – α) (x – β), where k is a constant
= k[x2 – (α + β)x + α β]
= kx2 – k(α + β)x + k α β
Comparing the coefficients of x2, x and constant terms on both the
sides, we get
a = k, b = – k(α + β) and c = kαβ.
This gives α + β =-b/a
αβ =c/a

10. EX-Find the zeroes of the polynomial x2 – 3 and verify the
relationship between the zeroes and the coefficients.
Recall the identity a2 – b2 = (a – b)(a + b). Using it, we can write:
x2 – 3 = ( x+√3 )(x− √3)
So, the value of x2 – 3 is zero when x = 3 or x = – 3⋅
Therefore, the zeroes of x2 –3 are x+ √3 and x-√3
Therefore
X= -√3=α
X=+√3=β

11. Now compare the f(x) to ax2 + bx + c
a=1 , b=0 , c=-3
Now we know that
α + β =-b/a
so √3+(-√3)=-0/1
0=0 (hence proved)
αβ =c/a
So √3 X (-√3) = -3/1
-3=-3(hence proved)