Zeros, also known as roots, of a polynomial are the values of x for which the polynomial equation P(x)=0 holds true. These values are crucial in various applications, from solving equations to analyzing the behaviour of polynomial functions.
Finding the zeros of a polynomial can involve different methods, such as factoring, using the Rational Root Theorem, synthetic division, or applying numerical techniques for more complex polynomials.
In this article, we will discuss different practice problems on the concept of zeros of polynomials.
What is Polynomial?
Polynomial is a mathematical expression consisting of variables, coefficients, and exponents combined using addition, subtraction, and multiplication operations. The exponents in a polynomial are whole numbers (non-negative integers), and the coefficients are typically real numbers, though they can also be complex numbers.
A polynomial in one variable x is generally written as:
P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0
Where:
- an, an-1, . . ., a1, a0 are coefficients.
- n is a non-negative integer representing the degree of the polynomial (the highest power of the variable).
- an is the leading coefficient, and a0 is the constant term.
Zeros of Polynomial
Zeros of a polynomial, also known as the roots of the polynomial, are the values of the variable that make the polynomial equal to zero. In other words, if P(x) is a polynomial, then x=r is a zero of the polynomial if P(r)=0.
Important Results Related to Zeros of Polynomial
- Remainder Theorem states that when you divide a polynomial P(x) by a linear polynomial x -a, the remainder is equal to P(a).
- Factor Theorem states that for any root of a polynomial i.e., x = a then (x - a) is the factor of a given polynomial.
- Rational Root Theorem states that any rational zero of the polynomial
P(x) = a_nx^n + \ldots + a_1x + a_0 is of the form\pm \frac{p}{q} , where p is a factor of the constant term a0 and q is a factor of the leading coefficient an.
Solved Examples on Zeros of Polynomial
Problem 1: Find the zeros of the polynomial p(x) = x2 - 5x + 6.
Solution:
p(x) = x2 - 5x + 6
⇒ x2 - 5x + 6 = 0
⇒ (x - 2)(x - 3) = 0
Either x - 2 = 0 or x - 3 = 0
⇒ x = 2 or x = 3
So, the zeros of the polynomial p(x) = x2 - 5x + 6 are x = 2 and x = 3.
Problem 2: Find the zeros of the polynomial p(x) = x3 - 4x2 + x + 6.
Solution:
p(x) = x3 - 4x2 + x + 6
Possible rational zeros are ± 1, ± 2, ± 3, ± 6.
For x = 1:
p(1) = 13 - 4(1)2 + 1 + 6 = 1 - 4 + 1 + 6 = 4 (not a zero)
For x = 2:
p(2) = 23 - 4(2)2 + 2 + 6 = 8 - 16 + 2 + 6 = 0 (not a zero)
Using synthetic division or polynomial division:
x3 - 4x2 + x + 6 = (x - 2)(x2 - 2x - 3)
As x2 - 2x - 3 = (x - 3)(x + 1)
Thus, x3 - 4x2 + x + 6 = (x - 2)(x - 3)(x + 1)
x - 2 = 0 or x - 3 = 0 or x + 1 = 0
x = 2, or x = 3 or x = -1
So, the zeros of the polynomial p(x) = x3 - 4x2 + x + 6 are x = 2, 3, -1.
Problem 3: Find the zeros of the polynomial p(x) = x4 - 5x3 + 6x2.
Solution:
p(x) = x4 - 5x3 + 6x2
⇒ p(x) = x2(x2 - 5x + 6)
As x2 - 5x + 6 = (x - 2)(x - 3)
Thus, p(x) = x2(x - 2)(x - 3)
Now, x2 = 0 or x - 2 = 0 or x - 3 = 0
x = 0 (multiplicity 2) or x = 2 or x = 3
So, the zeros of the polynomial p(x) = x4 - 5x3 + 6x2 are x = 0 (with multiplicity 2), x = 2, and x = 3.
Example 4: Solve P(x) = x3 - 6x3 - 6x + 36 by factoring the polynomial and identifying its zeros.
Solution:
P(x) = x3 - 6x3 - 6x + 36
Notice that we can group and factor by grouping:
P(x) = (x3 - 6x3 ) - (6x - 36)
Factor out the common terms:
= x2(x - 6) - 6(x - 6)
= (x - 6)(x2 - 6)
The zeros of the polynomial are the values of x for which P(x) = 0:
(x - 6)(x2 - 6) = 0
Either x - 6 = 0 ⇒ x = 6
Or x2 - 6 = 0 ⇒ x = ±√6
Thus, the zeros are 6, √6, -√6.
Problem 5: Determine whether x = 2 is a zero of P(x) = 2x3 + 3x2 - 5x + 6 using synthetic division, and find the remaining zeros.
Solution:
Synthetic division to check x = 2:
Set up the synthetic division:
\begin{array}{r|rrrr} 2 & 2 & 3 & -5 & 6 \\ & & 4 & 14 & 18 \\ \hline & 2 & 7 & 9 & 24 \end{array}
The remainder is 24, which means x = 2 is not a zero of the polynomial.
Problem 6: Find all possible rational zeros of the polynomial P(x) = 2x3 - 3x2 - 8x + 12 and verify which ones are actual zeros.
Solution:
For P(x) = 2x3 - 3x2 - 8x + 12
Factors of 12 (constant term): ± 1, ± 2, ± 3, ± 4, ± 6, ± 12
Factors of 2 (leading coefficient): ± 1, ± 2
Possible rational zeros: ± 1, ± 1/2, ± 2, ± 3, ± 3/2, ± 4, ± 6, ± 12
Evaluate P(x) for each possible rational zero.
For x = 1:
P(1) = 2(1)3 - 3(1)2 - 8(1) + 12 = 2 - 3 - 8 + 12 = 3 ≠ 0
For x = -1:
P(-1) = 2(-1)3 - 3(-1)2 - 8(-1) + 12 = -2 - 3 + 8 + 12 = 15 ≠ 0
For x = 2:
P(2) = 2(2)3 - 3(2)2 - 8(2) + 12 = 16 - 12 - 16 + 12 = 0
Thus, x = 2 is a zero of the polynomial P(x).
After identifying x = 2 as a zero, you can factor the polynomial using synthetic division or long division to find other zeros if necessary.
\begin{array}{r|rrrr} 2 & 2 & -3 & -8 & 12 \\ & & 4 & 2 & -12 \\ \hline & 2 & 1 & -6 & 0 \end{array} This gives us 2x3 - 3x2 - 8x + 12 = (x - 2)(2x2 + x - 6).
Factor 2x2 + x - 6 further:
2x2 + x - 6 = (2x - 3)(x + 2)
So, the polynomial can be factored as:
P(x) = (x - 2)(2x - 3)(x + 2)
The zeros of the polynomial P(x) = 2x3 - 3x2 - 8x + 12 are x = 2, 3/2, -2.
Zeros of Polynomial: Worksheet
Find the zeroes of following polynomials
- f(x) = x2 − 7x + 10
- g(x) = 2x3 − 3x2 − 8x + 12
- h(x) = x4 - 16
- a(x) = x3 + x2 − 4x - 4
- b(x) = 3x2 − 12x + 9
- c(x) = x3 - 9x
- d(x) = 4x4 - 20x2 + 25
- e(x) = x2 + 6x + 9
- i(x) = 2x3 − 5x2 − 4x + 3
- j(x) = x4 + x3 − 6x2
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