Polynomials

The word polynomial is derived from the Greek words ‘poly’ means ‘many‘ and ‘nominal’ means ‘terms‘, so altogether it said “many terms”. A polynomial can have any number of terms but not infinite. Learn about degree, terms, types, properties, polynomial functions in this article.

Examples: x – 2, 4y + 89, 3x – z.

Examples: x² – 2x + 5, x² – 3x etc.

Monomial: A monomial is an expression which contains only one term. For an expression to be a monomial, the single term should be a non-zero term. A few examples of monomials are:

Binomial: A binomial is a polynomial expression which contains exactly two terms. A binomial can be considered as a sum or difference between two or more monomials. A few examples of binomials are:

Trinomial

A trinomial is an expression which is composed of exactly three terms. A few examples of trinomial expressions are:

Step 1: To obtain the first term of the quotient, divide the highest degree term of the dividend by the highest degree term of the divisor. Then carry out the division process.

Step 2: To obtain the second term of the quotient, divide the highest degree term of the new dividend by the highest degree term of the divisor. Then again carry out the division process

Step 3: Continue the process till the degree of the new dividend is less that the degree of the divisor. This will be called the remainder.

If f(x) = g(x) q(x) + r(x) and r(x) = 0, then polynomial g(x) is a factor of the polynomial f(x).

Long Questions :

x³ – 4x² + 5x – 2; 2,1,1

Case Study Questions:

Assertion: If the sum of the zeroes of the quadratic polynomial x²-2kx+8 is 2 then value of k is 1.
Reason: Sum of zeroes of a quadratic polynomial ax²+bx+c is -b/a

deg p(x) < deg g(x)

Product of zeros = – 3 x 4 = -12

∴ Required polynomial = x² – x – 12

Then α + 6 = 5 3 ⇒ α = -1

Then α + (-α) = 0 ⇒ $-\frac{b}{a}$ = 0 ⇒ b = 0.

Then f(3) = 3² – 5 x 3 + 4 = -2

For f(3) = 0, 2 must be added to f(x).

Then the other root = -α

Sum of the roots = (-α) + α = 0

$\Rightarrow -\frac{b}{a}=0or-\frac{8k}{4}=0ork=0$

∴ (k – 1)(-3)² + k(-3) + 1 = 0 :

⇒ 9k – 9 – 3k + 1 = 0 ⇒ k = 4/3.

∴ p(1) = a(1)² – 3(a – 1) x 1 – 1 = 0

⇒ a – 3a + 3 – 1 = 0 ⇒ 2a = -2 ⇒ a = 1

= α + β – 3αβ = 5 – 3 x 6 = -13

For zeros, p(x) = 0

⇒ (2x – 3)(2x – 3) = 0 ⇒ x = $\frac{3}{2}$

: 8x⁴ + 14x³ – 2x² + 7x – 8 – y is exactly divisible by g(x)

Now,

∵ Remainder should be 0.

∴ 14x – 10 – y = 0 or 14x – 10 = y or y = 14x – 10

∴ (14x – 10) should be subtracted from p(x) so that it will be exactly divisible by g(x)

f(x) = g(x) × q(x) + r(x)

= f(x) – r(x) = g(x) × q(x) ⇒ f(x) + {-r(x)} = g(x) × q(x)

Clearly, RHS is divisible by g(x). Therefore, LHS is also divisible by g(x). Thus, if we add –r(x) to f(x), then the resulting polynomial is divisible by g(x). Let us now find the remainder when f(x) is divided by g(x).

∴ r(x) = -61x + 65 or -r(x) = 61x – 65

Hence, we should add –r(x) = 61x – 65 to f(x) so that the resulting polynomial is divisible by g(x).

On comparing with general polynomial px) ax³ + bx² + cx + d, we get a = 1, b = -4, c = 5 and d = -2

Given zeros 2, 1, 1.

∴ p(2) = (2)³ – 4(2)² + 5(2) – 2 = 8 – 16 + 10 – 2 = 0

and p(1) = (1)³ – 4(1)² + 5(1) – 2 = 1 – 4 + 5 – 2 = 0

Hence, 2, 1 and I are the zeros of the given cubic polynomial.

Again, consider α = 2, β = 1, γ = 1

∴ α + 13 + y = 2 + 1 + 1 = 4

p(x) = a[x³ + (-2)x² + (-7)x + 14] ⇒ p(x) = a[x³ – 2x² – 7x + 14]

For real value of a = 1, p(x) = x³ – 2x² – 7x + 14

Now, α + β + γ = 5 α + β – 2 = 5

= α + β = 7 a = 7 – β

= (7 – β) β =12 ⇒ 7β – β² – 12

= β² + 7β + 12 = 0 ⇒ β² – 3β – 4β + 12 = O

= β = 4 or β = 3

β = 4 or β = 3

∴ α = 3 or α = 4

On dividing f(x) by (x – √2), we get

∴ α, β and γ are zeros of the polynomial p(x).

β – α = γ + β ⇒ 2β = α + γ

α + β + γ = $-\frac{b}{a}$ = $\frac{-(-12)}{1}$ = 12 ⇒ α + γ = 12 – β …….. (i)

From (i) and (ii)

2β = 12 – β or 3β = 12 or β = 4

Putting the value of β in (i), we have

8 = a + γ

αβγ = $–\frac{d}{a}=\frac{-(-28)}{1}=28$ …….. (iii)

(αγ) 4 = 28 or αγ = 7 or γ = 7α ….. (iv)

Putting the value of γ = 7α in (iii), we get

⇒ 8 = α + $\frac{7}{\alpha }$ ⇒ 8α = α² + 7

⇒ α² – 8α + 7 = 0 ⇒ α² – 7α – 1α + 7 = 0

⇒ α(α – 7)-1 (α – 7) = 0 ⇒ (α – 1)(α – 7) = 0

⇒ α = 1 or α = 7

Solution:

Since, the given graph is parabolic is shape, therefore it will represent a quadratic polynomial.

[∵ Graph of quadratic polynomial is parabolic in shape)

Solution:

Since, the graph cuts the x-axis at -1, 5. So the polynomial has 2 zeroes i.e., -1 and 5.

Solution:

Sum of zeroes = -1 + 5 = 4

Solution:

Since α and β are zeroes of the given polynomial and β > α, β > α,

∴ α = -1 ∴ α = –1 and β = 5β = 5

∴ |8α + β| = |8(-1) + 5| = | -8 + 5| = |-3| = 3.

∴|8α + β|=|8(−1)+5|=|−8+5|=|−3|=3.

Solution:

Since the zeroes of the given polynomial are -1 and 5.

∴ Required polynomial p(x)

= k² {x² – (-1 + 5)x + (-1)(5)} = k(x² – 4x – 5)

For k = -1, we get,

p(x) = -x² + 4x + 5, which is the required polynomial.

Solution:

Graph of a quadratic polynomial is a parabolic in shape.

Solution:

Since the graph of the polynomial cuts the x-axis at (-6, 0) and ( 6, 0). So, the zeroes of polynomial are -6 and 6.

∴ Required polynomial is

p(x) = x² – (-6 + 6)x + (-6)(6) = x² – 36

Solution:

We have, p(x) = x² – 36

Now, p(6) = 6² – 36 = 36 – 36 = 0

Solution:

Let f(x) = x² + 2x – 3. Then,

Solution:

The given polynomial is at² + 5t + 3a

Given, sum of zeroes = product of zeroes