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**What is a Polynomial?Answer:**

Polynomial p(x) in one variable x is an algebraic expression in x of the form

p(x) = a_{n}x^{n} + a_{n–1}x^{n – 1} + . . . + a_{2}x^{2} + a_{1}x + a_{0},

where a_{0}, a_{1}, a_{2}, . . ., an are constants and a_{n} ≠ 0.

a_{0}, a_{1}, a_{2}, . . ., an are respectively the coefficients of x^{0}, x, x^{2}, . . ., x^{n},

and n is called the degree of the polynomial.

Each of a_{n}x^{n}, a_{n–1} x^{n–1}, ..., a_{0}, with a_{n} ≠ 0, is called a term of the polynomial p(x).

**Type of PolynomialAnswer:**

A polynomial of one term is called a monomial.

Example p(x) = x

A polynomial of two terms is called a binomial.

Example q(x) = x^{2} – x,

A polynomial of three terms is called a trinomial.

Examples r(y) = y^{3} + y + 1

A polynomial of degree one is called a linear polynomial.

Example p(x) = x +1

A polynomial of degree two is called a quadratic polynomial.

Examples p(x) = x^{2} + x + 2

A polynomial of degree three is called a cubic polynomial.

Example r(y) = y^{3} + y + 1

**Other Information **A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0. In this case, a is also called a root of the equation p(x) = 0.

Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial has no zero, and every real number is a zero of the zero polynomial.

**Remainder Theorem :** If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is ided by the linear polynomial x – a, then the remainder is p(a).

**Factor Theorem :** x – a is a factor of the polynomial p(x), if p(a) = 0. Also, if x – a is a factor of p(x), then p(a) = 0.

The highest power of the variable in a polynomial as the degree of the polynomial

The degree of a non-zero constant polynomial is zero.

**Identity I : **(x + y)^{2} = x^{2} + 2xy + y^{2}

**Identity II : **(x – y)^{2} = x^{2} – 2xy + y^{2}

**Identity III : **x^{2} – y^{2} = (x + y) (x – y)

**Identity IV : **(x + a) (x + b) = x^{2} + (a + b)x + ab**Identity V : **(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx**Identity VI : **(x + y)^{3} = x^{3} + y^{3} + 3xy (x + y)**Identity VII : **(x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)

= x^{3} – 3x^{2}y + 3xy^{2} – y^{3}**Identity VIII : **x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – zx)

**Question 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x ^{2} – 3x + 7 (ii) y^{2} + √2 (iii) 3√t + t√ 2 (iv) y +2/y (v) x^{10} + y^{3} + t^{50}**

**Solution (i)4x ^{2} – 3x + 7**

there is only one variable x with whole number power so this polynomial in one variable

**(ii)y ^{2} + √2 **

there is only one variable y with whole number power so this polynomial in one variable

**(iii) 3√t + t√ 2 **

there is only one variable t but in 3√t power of t is ½ which is not a whole number so 3√t + t√ 2 is not a polynomial

**(iv)y +2/y **

there is only one variable y but 2/y = 2y-1 so the power is not a whole number so y +2/y is not a polynomial

**(v) x ^{10} + y^{3} + t^{50} **

there are three variable x y and t and there powers are whole number so this polynomial in three variable

**Question 2. Write the coefficients of x2 in each of the following: (i) 2 + x ^{2} + x (ii) 2 – x^{2} + x^{3} (iii) (pi/2 )x^{2}+ x (iv) √2x – 1 Solution: (i) 2 + x^{2} + x **

Coefficient if x

**(ii) 2 – x ^{2} + x^{3} **

Coefficient of x

Coefficient of x

**(iv) √2x – 1 **

There are no x^{2} term so Coefficient of x^{2} = 0

**Question 3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.**

**Solution example of a binomial of degree 35**

degree is 35 so maximum power of term will 35 and it is binomial so it has 2 terms

so such example is x^{35} + 1 **example of a monomial of degree 100**

degree is 100 so maximum power of term will 100 and it is monomial so it has 1 terms

so such example is x^{100}

**Question 4. Write the degree of each of the following polynomials: (i) 5x ^{3} + 4x^{2} + 7x (ii) 4 – y^{2} (iii) 5t – √7 (iv) 3**

**Solution: (i)5x ^{3} + 4x^{2} + 7x**In above expression 5x

**(ii) 4 – y ^{2}**

in above expression –y2 has highest power, and power is 2 so this is the polynomial of degree 2

in above expression 5t has highest power, and power is 1 so this is the polynomial of degree 1

**(iv) 3 **

there is no variable therefore degree is 0

**Question 5. Classify the following as linear, quadratic and cubic polynomials: (i) x ^{2} + x (ii) x – x^{3} (iii) y + y^{2} + 4 (iv) 1 + x (v) 3t (vi) r^{2} **

**Solution: (i) x ^{2} + x **

Highest power of x is 2 so it is a quadratic polynomials

**(ii) x – x ^{3} **

Highest power of x is 3 so it is cubic polynomial

**(iii) y + y ^{2} + 4 **

Highest power of y is 2 so it is quadratic polynomial

**(iv) 1 + x**

Highest power of x is 1 so it is linear polynomial

**(v) 3t **

Highest power of t is 1 so it is linear polynomial

**(vi) r ^{2} **

Highest power of t is 2 so it is quadratic polynomial

**(vii) 7x ^{3}**

Highest power of x is 3 so it is cubic polynomial

**Question 1. Find the value of the polynomial 5x – 4x ^{2} + 3 at (i) x = 0 (ii) x = –1 (iii) x = 2**

**Solution: (i)x = 0**

5x – 4x^{2} + 3

Plug x = 0 we get

=>5(0) – 4(0)^{2} + 3

=>0 -0 + 3

=> 3

**(ii) x = –1 **

Plug x = - 1 we get

=>5x – 4x^{2} + 3

=>5(-1) – 4(-1)^{2} + 3

=> -5 – 4 +3

=>-6

**(iii) x = 2**

Plug x = 2 we get

=>5(2) – 4(2)^{2} + 3

=>10 – 16 + 3

=> -3

**Question 2. Find p(0), p(1) and p(2) for each of the following polynomials: (i) p(y) = y ^{2} – y + 1 (ii) p(t) = 2 + t + 2t^{2} – t3 (iii) p(x) = x3 (iv) p(x) = (x – 1) (x + 1)**

**Solution: (i)p(y) = y ^{2} – y + 1 **

Plug y = 0 we get

=>p(0) = (0)

=>p(0) = 0 – 0 + 1

=> 1

Plug y = 1 we get

=>p(1) = (1)^{2} – 1 + 1

=>p(1) = 1 – 1 + 1

=> 1

Plug y = 2 we get

=>p(2) = (2)^{2} – 2 + 1

=>p(2) = 4 – 2 + 1

=> 3

**(ii) p(t) = 2 + t + 2t ^{2} – t^{3} **

Plug t = 0 we get

=> p(t) = 2 + t + 2t

=> p(0) = 2 + 0 +2(0)

=> p(0) = 2 + 0 +0 – 0

=> p(0) = 2

p(t) = 2 + t + 2t^{2} – t^{3}

Plug t = 1 we get

=> p(t) = 2 + t + 2t^{2} – t^{3}

=> p(1) = 2 + 1 +2(1)^{2} – (1)^{3}

=> p(1) = 2 + 1 + 2 – 1

=> p(1) = 4

p(t) = 2 + t + 2t^{2} – t^{3}

Plug t = 2 we get

=> p(t) = 2 + t + 2t^{2} – t^{3}

=> p(2) = 2 + 2 +2(2)^{2} – (2)^{3}

=> p(2) = 2 + 2 + 8 – 8

=> p(2) = 4

**(iii) p(x) = x ^{3} **

=>plug x = 0 we get

=>p(x) = x

=>p(0) = (0)

=> p(0) = 0

p(x) = x3

=>plug x = 1 we get

=>p(x) = x^{3}

=>p(1) = (1)^{3}

=> p(1) = 1

p(x) = x^{3}

=>plug x = 2 we get

=>p(x) = x^{3}

=>p(2) = (2)^{3}

=> p(2) = 8

**(iv) p(x) = (x – 1) (x + 1)**

Plug x = 0 we get

=>p(x) = (x – 1) (x + 1)

=> p(0) = (0 – 1) (0 + 1)

=> p(0) = (- 1)(1)

=> p(0) = - 1

Plug x = 1 we get

=>p(x) = (x – 1) (x + 1)

=> p(1) = (1 – 1) (1 + 1)

=> p(1) = (0)(2)

=> p(1) = 0

Plug x = 2 we get

=>p(x) = (x – 1) (x + 1)

=> p(2) = (2 – 1) (2 + 1)

=> p(2) = (1)(3)

=> p(2) = 3

**Question 3. Verify whether the following are zeroes of the polynomial, indicated against them. (i) p(x) = 3x + 1, x = - 1/3 (ii) p(x) = 5x – π, x = 4/5 (iii) p(x) = x**

**Solution: (i) p(x) = 3x + 1, x = - 1/3**

Plug x = -1/3

=> p(x) = 3x + 1

=>p(-1/3) = 3(-1/3) +1

=>p(-1/3) = -1 +1

=>p(-1/3) = 0

When P(a) = 0 then a is always zero of polynomial

**Hence -1/3 is zero of polynomial p(x) = 3x + 1**

**(ii) p(x) = 5x – π, x = 4/5**

Plug x = 4/5 we get

=> p(4/5) = 5x – π,

=> p(4/5) = 5(4/5) – π,

=> p(4/5) = 4 – π,

And pi = 22/7 so that

=> p(4/5) = 4 – 22/7 is not = 0

**Hence 4/5 is not zero of polynomial p(x) = 5x – π**

**(iii) p(x) = x ^{2} – 1, x = 1, –1**

Plug x = - 1

=> p(x) = x

=> p(-1) = (-1)

=> p(-1) = 1 – 1

=> p(-1) = 0

Plug x = 1

=> p(x) = x^{2} – 1

=> p(1) = (1)^{2} – 1

=> p(1) = 1 – 1

=> p(1) = 0

**Hence both x = - 1 and 1 are zero of polynomial p(x) = x ^{2} – 1**

**(iv) p(x) = (x + 1) (x – 2), x = – 1, 2**

Plug x = - 1

=>p(x) = (x + 1) (x – 2)

=>p(-1) = (-1 + 1) (-1 – 2)

=>p(-1) = (0) (-3)

=>p(-1) = 0

Now plug x = 2 we get

Plug x = 2

=>p(x) = (x + 1) (x – 2)

=>p(2) = (2 + 1) (2 – 2)

=>p(2) = (3) (0)

=>p(2) = 0

**Hence -1 and 2 both are zero of the polynomial p(x) = (x + 1) (x – 2) **

**(v) p(x) = x ^{2}, x = 0 **

Plug x = 0 we get

=>p(x) = x

=>p(0) = (0)

=>p(0) = 0

**(vi) p(x) = lx + m, x = –m/l**

Plug x = - m/l we get

=> p(x) = lx + m

=> p(-m/l) = l(-m/l) + m

=> p(-m/l) = -m + m

=> p(-m/l) = 0

**Hence - m/l is the zero of polynomial p(x) = lx + m**

**(vii) p(x) = 3x ^{2} – 1, x = - 1/√3 , 2/√3**

Plug x = - 1/ √3 we get

=>p(x) = 3x^{2} – 1

=>p(-1/√3) = 3(-1/√3)^{2} – 1

=>p(-1/√3) = 3(1/3) – 1

=>p(-1/√3) = 1 – 1

=>p(-1/√3) = 0

Plug x = 2/ √3 we get

=>p(x) = 3x^{2} – 1

=>p(1/√3) = 3(2/√3)^{2} – 1

=>p(1/√3) = 3(4/3) – 1

=>p(1/√3) = 4 – 1

=>p(1/√3) = 3

**Hence x = - 1/√3 is zero of the polynomial p(x) = 3x ^{2} – 1**

**(viii) p(x) = 2x + 1, x =1/2**

Plug x = ½ we get

=> p(x) = 2x + 1

=> p(1/2) = 2(1/2) + 1

=> p(1/2) = 2(1/2) + 1

=> p(1/2) = 1 + 1

=> p(1/2) = 2

**Hence ½ is not a zero of polynomial p(x) = 2x + 1**

**Question 4. Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.**

**Solution: (i)p(x) = x + 5 **Plug p(x) = 0 we get

=>x+5 =0

=> x = - 5

**(ii) p(x) = x – 5 **

Plug p(x) = 0 we get

=>x - 5 = 0

=> x = 5

**5 is zero of the polynomial **

**(iii) p(x) = 2x + 5 **

Plug p(x) = 0 we get

=>2x+5 =0

=> 2x = - 5

=> x= -5/2

**-5/2is zero of the polynomial **

**(iv) p(x) = 3x – 2 **

Plug p(x) = 0 we get

=>3x-2=0

=>3x =2

=> x = 2/3

**2/3 is zero of the polynomial **

**(v) p(x) = 3x **

Plug p(x) = 0 we get

=>3x=0

=> x= 0/3

=> x=0

**0 is zero of the polynomial **

**(vi) p(x) = ax, a ≠ 0 **

Plug p(x) = 0 we get

=>ax=0

=> x =0/a

=> x = 0

**0 is zero of the polynomial **

**(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.**

Plug p(x) = 0 we get

=>cx +d = 0

=> cx = - d

=> x = -d/c

**-d/c is zero of the polynomial **

**Question 1. Find the remainder when x ^{3}+3x^{2} + 3x + 1 is ided by**

(i) x + 1 (ii) x –1/2 (iii) x (iv) x + π (v) 5 + 2x

**Solution:(i) x + 1 **

Apply remainder theorem

=>x + 1 =0

=> x = - 1

Replace x by – 1 we get

=>x^{3}+3x^{2} + 3x + 1

=>(-1)^{3} + 3(-1)^{2} + 3(-1) + 1

=> -1 + 3 - 3 + 1

=> 0

Remainder is 0

**(ii) x –1/2 **

Apply remainder theorem

=>x – 1/2 =0

=> x = 1/2

Replace x by 1/2 we get

=>x^{3}+3x^{2} + 3x + 1

=>(1/2)^{3} + 3(1/2)^{2} + 3(1/2) + 1

=> 1/8 + 3/4 + 3/2 + 1

Add the fraction taking LCM of denominator we get

=>(1 + 6 + 12 + 8)/8

=>27/8

Remainder is 27/8

**(iii) x **

Apply remainder theorem

=>x =0

Replace x by 0 we get

=>x^{3}+3x^{2} + 3x + 1

=>(0)^{3} + 3(0)^{2} + 3(0) + 1

=> 0+0 +0 + 1

=> 1

Remainder is 1

**(iv) x + π **

Apply remainder theorem

=>x + π =0

=> x = - π

Replace x by – π we get

=>x^{3}+3x^{2} + 3x + 1

=>(- π)^{3} + 3(-π)^{2} + 3(-π) + 1

=> - π3 + 3π2 - 3π + 1

Remainder is - π^{3} + 3π^{2} - 3π + 1

**(v) 5 + 2x**

Apply remainder theorem

=>5+2x =0

=> 2x = - 5

=> x = - 5/2

Replace x by – 5/2 we get

=>x^{3}+3x^{2} + 3x + 1

=>(-5/2)^{3} + 3(-5/2)^{2} + 3(-5/2) + 1

=> -125/8 + 75/4 – 15/2 + 1

Add the fraction taking LCM of denominator

=>(-125 + 150 - 60 + 8 )/125

=> -27/8

Remainder is -27/8

**Question 2. Find the remainder when x ^{3} – ax^{2} + 6x – a is ided by x – a.** Apply remainder theorem

Solution:

=>x – a =0

=> x = a

Replace x by a we get

=> x

=>( a)

=> a

=> 5a

**Question 3. Check whether 7 + 3x is a factor of 3x ^{3} + 7xSolution:** Apply remainder theorem

=>7 + 3x =0

=> 3x = - 7

=> x = - 7/3

Replace x by - 7/3 we get

=>3x

=>3(-7/3)

=>3(-343/27) – 49/3

=> (-343/9) – 49/3

This value is not equal to 0

**Question 1. Determine which of the following polynomials has (x + 1) a factor : (i) x ^{3} + x^{3} + x + 1 (ii) x^{4} + x^{3} + x^{3} + x + 1 (iii) x^{4} + 3x^{3} + 3x^{3} + x + 1 (iv) x^{3} – x^{3} – (2+√2)x + √2**

**Solution: (i) x ^{3} + x^{3} + x + 1 **Apply remainder theorem

=>x + 1 =0

=> x = - 1

Replace x by – 1 we get

=>x

=>(-1)

=> -1 + 1 - 1 + 1

=> 0

**(ii)x ^{4} + x^{3} + x^{3} + x + 1**

Apply remainder theorem

=>x + 1 =0

=> x = - 1

Replace x by – 1 we get

=>x

=> (-1)

=> 1 -1 + 1 - 1 + 1

=> 1

Apply remainder theorem

=>x + 1 =0

=> x = - 1

Replace x by – 1 we get

=>x

=> (-1)

=> 1 -3 + 3 - 1 + 1

=> 1

Apply remainder theorem

=>x + 1 =0

=> x = - 1

Replace x by – 1 we get

=>x

=> (-1)

=> 1 - 1 + 2 + √2 + √2

=> 2 + 2√2

**Question 2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x**

**Solution: (i) p(x) = 2x ^{3} + x^{2} – 2x – 1, g(x) = x + 1 **

Apply remainder theorem

=>x + 1 =0

=> x = - 1

Replace x by – 1 we get

=>2x

=>2(-1)

=> -2 + 1 + 2 - 1

=> 0

**(ii) p(x) = x ^{3} + 3x^{2} + 3x + 1, g(x) = x + 2**

Apply remainder theorem

=>x + 2 =0

=> x = - 2

Replace x by – 2 we get

=>x

=>(-2)

=> -8 + 12 - 6 + 1

=> -1

**(iii) p(x) = x ^{3} – 4x^{2} + x + 6, g(x) = x – 3**

Apply remainder theorem

=>x - 3 =0

=> x = 3

Replace x by – 2 we get

=>x

=>(3)

=> 27 - 36 +3 + 6

=> 0

**Question 3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases: (i) p(x) = x ^{2} + x + k (ii) p(x) = 2x^{2} + kx + √2 (iii) p(x) = kx^{2} – 2x + 1 (iv) p(x) = kx^{2} – 3x + k**

**Solution: (i) p(x) = x ^{2} + x + k **

Apply remainder theorem

=>x - 1 =0

=> x = 1

According to remainder theorem p(1) = 0 we get

Plug x = 1 we get

=> k(1)

=>k +1 + 1 =0

=> k + 2 = 0

=> k = - 2

**(ii) p(x) = 2x ^{2} + kx + √2**

Apply remainder theorem

=>x - 1 =0

=> x = 1

According to remainder theorem p(1) = 0 we get

Plug x = 1 we get

p(1) = 2(1)

p(1) =2 + k + √2

0 = 2 + √2 + k

-2 - √2 = k

- (2 + √2) = k

**(iii) p(x) = kx2 – √2x + 1 **

Apply remainder theorem

=>x - 1 =0

=> x = 1

According to remainder theorem p(1) = 0 we get

Plug x = 1 we get

p(1) = k(1)^{2} – √2(1)+ 1

P(1) = K - √2 + 1

0 = K - √2 + 1

√2 -1 = K

**Answer k= √2 -1 (iv) p(x) = kx**Apply remainder theorem

=>x - 1 =0

=> x = 1

According to remainder theorem p(1) = 0 we get

Plug x = 1 we get

P(1) = k(1)

0= k – 3 + k

0 = 2k – 3

3 = 2k

3/2 = k

**Answer k = 3/2 **

**Question 4. Factorise : (i) 12x ^{2} – 7x + 1 (ii) 2x^{2} + 7x + 3 (iii) 6x^{2} + 5x – 6 (iv) 3x^{2} – x – 4**

**Solution (i) 12x ^{2} – 7x + 1**

**Questoin 5. Factorise : (i) x ^{3} - 2x^{2} - x + 2 (ii) x^{3} - 3x^{2} - 9x - 5 (iii) x^{3} + 13x^{2} + 32x + 20 (iv) 2y^{3} + y^{2} - 2y - 1 **

**Solution:(i) x ^{3} - 2x^{2} - x + 2 **

Solution (i) Let take f(x) = x3 - 2x2 - x + 2

The constant term in f(x) is are ±1 and ±2

Putting x = 1 in f(x), we have

f(1) = (1)

= 1 - 2 - 1 + 2 = 0

According to remainder theorem f(1) = 0 so that (x - 1) is a factor of x3 - 2x2 - x + 2

Putting x = - 1 in f(x), we have

f(-1) = (-1)

= -1 - 2 + 1 + 2 = 0

According to remainder theorem f(-1) = 0 so that (x + 1) is a factor of x3 - 2x2 - x + 2

Putting x = 2 in f(x), we have

f(2) = (2)

= 8 -82 - 2 + 2 = 0

According to remainder theorem f(2) = 0 so that (x – 2 ) is a factor of x3 - 2x2 - x + 2

Here maximum power of x is 3 so that its can have maximum 3 factors

**So our answer is (x-1)(x+1)(x-2)**

**(ii) x ^{3} - 3x^{2} - 9x - 5 **

Possible zeros are factors of ± constant term / coefficient of leading term

Here constant term is -5 and coefficient of leading term is 1

So that possible zeros will ±1 and ±5

Take f(x) = x

Plug x = 1 we get

f(1) = (1)

f(1) = 1 – 3 - 9 – 5

F(1) = -16 ≠ 0

So that (x-1) is not a factor of x

Plug x = -1

f(-1) = (-1)

f(1) = -1 – 3 + 9 – 5

F(1) = 0 so it a zero

x-1 = 0

x+1 = 0

So that (x+1) is a factor of x

Divide the expression by x+1 we get

Here coefficient of leading term is 1 and constant term is 20

So possible zeros are factors of ± 20/1

So possible zeros are ±1 ,±2,±4,±5,±10, and ±20

And here all terms are positive so that zeros cannot positive

Plug x = -1

=>x

=> (-1)

=> -1+ 13 - 32 + 20

=> 0

So that (x+1) is a factor x

Plug x = - 2

=>x^{3} + 13x^{2} + 32x + 20

=> (-2)^{3} + 13(-2)^{2} + 32(-2) + 20

=> -8+ 52 - 64 + 20

=> 0

So that (x+2) is a factor x^{3} + 13x^{2} + 32x + 20

As we have already find two zeros third zeros can 20 / 1*2 = 10

Plug x = 10 we get

Plug x = - 2

=>x^{3} + 13x^{2} + 32x + 20

=> (-10)^{3} + 13(-10)^{2} + 32(-10) + 20

=> -1000+ 1300 - 320 + 20

=> 0

So that (x+10) is a factor x^{3} + 13x^{2} + 32x + 20

As leading term has 3 powers so that there are only 3 roots are possible

**Answer (x+1)(x+2)(x+10)**

**(iv) 2y ^{3} + y^{2} - 2y – 1**

Here constant term is -1

Coefficient of leading term is 2

So possible zeros are ±1 ,±1/2

Plug y = 1

=>2y

=>2(1)

=> 2+ 1 -2-1

=>0

Here y= -1

Y+1 =0 so that

(y+1) is factor of 2y

Plug y = -1

=>2y^{3} + y^{2} - 2y – 1

=>2(-1)^{3} + (-1)^{2} – 2(-1) – 1

=> - 2+ 1 + 2-1

=>0

Here y= 1

y - 1 =0 so that

(y - 1) is factor of 2y^{3} + y^{2} - 2y – 1

Plug y = ½

=>2y^{3} + y^{2} - 2y – 1

=>2(½)^{3} + (½)^{2} – 2(½) – 1

=> 2/8+ 1/4 -1-1

=>-3/2 ≠ 0

(y – ½) is factor of 2y^{3} + y^{2} - 2y – 1

Plug y = -½

=>2y^{3} + y^{2} - 2y – 1

=>2(-½)^{3} + (-½)^{2} – 2(-½) – 1

=> -2/8+ ¼ + 1-1

=>0

(y + ½) is factor of 2y^{3} + y^{2} - 2y – 1

Here y has max powers 3 so there are 3 possible factors

**And our answer is (y-1)(y+1)(y+ -½)**

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