How many unique roots will a third degree polynomial function have?

How many unique roots will a third degree polynomial function have?
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1 Answer

Answer :

It can have 1, 2 or 3 unique roots.
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Last Answer : (a) Let p (x) = 5x – 4x2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6

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Description : Zero of the polynomial p(x)=2x+5 is -Maths 9th

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x + 1 is a factor of the polynomial -Maths 9th

Description : x + 1 is a factor of the polynomial -Maths 9th

Last Answer : (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. So, x = -1 is zero of x3 + x2 + x+1 (-1)3 + (-1)2 + (-1) + 1 = 0 ⇒ -1+1-1 + 1 = 0 ⇒ 0 = 0 Hence, our assumption is true.

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Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th

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Last Answer : Let p(x) =3x3 – 4x2 + 7x – 5 At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3)- 5 = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively.

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Last Answer : Given, polynomial is p(x) = (x – 2)2 – (x+ 2)2 For zeroes of polynomial, put p(x) = 0 (x – 2)2 – (x+ 2)2 = 0 (x-2 + x+2)(x-2-x-2) = 0 [using identity, a2-b2 =(a-b)(a + b)] ⇒ (2x)(-4) = 0

By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial x4 + 1 and x-1. -Maths 9th

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Last Answer : Actual division method

Determine which of the following polynomial has x – 2 a factor -Maths 9th

Description : Determine which of the following polynomial has x – 2 a factor -Maths 9th

Last Answer : first option is the correct answer for the given question solution is as follows:- let x-2=0 then, x=2 put x in (i) 3(2)(2)+6(2)-24=0 12+12-24=0 {use BODMAS rule for solution}... 24-24=0 0=0 this verifies our answer

The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Description : The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Last Answer : p(x) is divided by x+ 2 =

Determine the remainder when polynomial p(x) is divided by x - 2 . -Maths 9th

Description : Determine the remainder when polynomial p(x) is divided by x - 2 . -Maths 9th

Last Answer : p(x) = x4 - 3x2 + 2x - 5 According to remainder theorem, the required remainder will be = p(2) p(x) = x4 - 3x2 + 2x - 5 ∴ p(2) = 24 - 3(2)2 + 2(2) - 5 =16 - 12 + 4 - 5 = 3

Which one of the following is a polynomial ? -Maths 9th

Description : Which one of the following is a polynomial ? -Maths 9th

Last Answer : Following is a Polynomial

The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Description : The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Last Answer : (a) Let p (x) = 5x – 4x2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6

Zero of the zero polynomial is -Maths 9th

Description : Zero of the zero polynomial is -Maths 9th

Last Answer : (c) Zero of the zero polynomial is any real number. e.g., Let us consider zero polynomial be 0(x-k), where k is a real number For determining the zero, put x-k = 0 ⇒ x = k Hence, zero of the zero polynomial be any real number.

Zero of the polynomial p(x)=2x+5 is -Maths 9th

Description : Zero of the polynomial p(x)=2x+5 is -Maths 9th

Last Answer : (b) Given, p(x) = 2x+5 For zero of the polynomial, put p(x) = 0 ∴ 2x + 5 = 0 ⇒ -5/2 Hence, zero of the polynomial p(x) is -5/2.

One of the zeroes of the polynomial 2x2 + 7x – 4 is -Maths 9th

Description : One of the zeroes of the polynomial 2x2 + 7x – 4 is -Maths 9th

Last Answer : (b) Let p (x) = 2x2 + 7x-4 = 2x2 + 8x-x-4 [by splitting middle term] = 2x(x+ 4)-1(x+ 4) = (2x-1)(x+ 4) For zeroes of p(x), put p(x) = 0 ⇒ (2x -1) (x + 4) = 0 ⇒ 2x-1 = 0 and x+4 = 0 ⇒ x = 1/2 and x = -4 Hence, one of the zeroes of the polynomial p(x) is 1/2.

If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th

Description : If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th

Last Answer : (c) Let p(x) = 2x2 + kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 ⇒ 2-k = 0 ⇒ k= 2 Hence, the value of k is 2.

x + 1 is a factor of the polynomial -Maths 9th

Description : x + 1 is a factor of the polynomial -Maths 9th

Last Answer : (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. So, x = -1 is zero of x3 + x2 + x+1 (-1)3 + (-1)2 + (-1) + 1 = 0 ⇒ -1+1-1 + 1 = 0 ⇒ 0 = 0 Hence, our assumption is true.

Give an example of a polynomial, which is -Maths 9th

Description : Give an example of a polynomial, which is -Maths 9th

Last Answer : (i) The example of monomial of degree 1 is 5y or 10x. (ii) The example of binomial of degree 20 is 6x20 + x11 or x20 +1 (iii) The example of trinomial of degree 2 is x2 – 5x+ 4 or 2x2 -x-1

Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th

Description : Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th

Last Answer : Let p(x) =3x3 – 4x2 + 7x – 5 At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3)- 5 = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively.

Find the zeroes of the polynomial in each of the following, -Maths 9th

Description : Find the zeroes of the polynomial in each of the following, -Maths 9th

Last Answer : (i) Given, polynomial is p(x) = x- 4 For zero of polynomial, put p(x) = x-4 = 0 ⇒ x= 4 Hence, zero of polynomial is 4. (ii) Given, polynomial is g(x) = 3-6x For zero of polynomial, put g(x) ... polynomial h(y) = 2 y For zero of polynomial, put h(y) = 0 2y= 0 Hence, the zero of polynomial is 0,

Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. -Maths 9th

Description : Find the zeroes of the polynomial p(x)= (x – 2)2 – (x+ 2)2. -Maths 9th

Last Answer : Given, polynomial is p(x) = (x – 2)2 – (x+ 2)2 For zeroes of polynomial, put p(x) = 0 (x – 2)2 – (x+ 2)2 = 0 (x-2 + x+2)(x-2-x-2) = 0 [using identity, a2-b2 =(a-b)(a + b)] ⇒ (2x)(-4) = 0

By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial x4 + 1 and x-1. -Maths 9th

Description : By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial x4 + 1 and x-1. -Maths 9th

Last Answer : Actual division method

Determine which of the following polynomial has x – 2 a factor -Maths 9th

Description : Determine which of the following polynomial has x – 2 a factor -Maths 9th

Last Answer : first option is the correct answer for the given question solution is as follows:- let x-2=0 then, x=2 put x in (i) 3(2)(2)+6(2)-24=0 12+12-24=0 {use BODMAS rule for solution}... 24-24=0 0=0 this verifies our answer

The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Description : The polynomial p{x = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. -Maths 9th

Last Answer : p(x) is divided by x+ 2 =

FOR THE POLYNOMIAL -Maths 9th

Description : FOR THE POLYNOMIAL -Maths 9th

Last Answer : NEED ANSWER

FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Last Answer : As the given polynomial divisible by x-2 means the polynomial satisfies for the value x=2 So putting x=2 in x²+(4-k)x+2 yields 0 ⇒2²+(4-k)2+2=0 ⇒4+8-2k+2=0 ⇒ 2k=14 ⇒ k= ... ;-3x+2 if factorized yields (x-1)(x-2). Thus is divisible by x-2 as well as divisible by x-1.

FOR THE POLYNOMIAL -Maths 9th

Description : FOR THE POLYNOMIAL -Maths 9th

Last Answer : This answer was deleted by our moderators...