Polynomials ⇒⇒ Exercise 2.3

Question 1 (i)

Determine which of the following polynomials has (x + 1) a factor :
(i) x³ + x² + x + 1

Solution :

Let P(x) = x³ + x² + x + 1

To find the zero of the ( x + 1), we need to equate it to zero.

x + 1 = 0

x = -1

Given , P(x) = x³ + 3x² + 3x + 1

If (x + 1) is a factor of p (x) = x³ + x² + x + 1 , p (-1) must be zero

Substituting the value of x in given expression, we get;

P(-1) = (-1)³ + (-1)² + (-1) + 1

P(-1) = -1 + 1 - 1 + 1

P(-1) =0

Since, p(–1) = 0

Therefore, according to Factor Theorem (x + 1) is a factor of given polynomial.

Question 1 (ii)

Determine which of the following polynomials has (x + 1) a factor :
(ii) x⁴ + x³ + x² + x + 1

Solution :

Let P(x) = x⁴ + x³ + x² + x + 1

To find the zero of the ( x + 1), we need to equate it to zero.

x + 1 = 0

x = -1

Given , P(x) = x⁴ + x³ + x² + x + 1

If (x + 1) is a factor of p (x) = x⁴ + x³ + x² + x + 1 , p (-1) must be zero

Substituting the value of x in given expression, we get;

P(-1) = (-1)⁴ +(-1)³ + (-1)² + (-1) + 1

P(-1) = 1 -1 + 1 - 1 + 1

P(-1) = 1

Since, p(–1) ${ \ne }$ 0

Therefore, according to Factor Theorem (x + 1) is not a factor of given polynomial.

Question 1 (iii)

Determine which of the following polynomials has (x + 1) a factor :
(iii) x⁴ + 3x³ + 3x² + x + 1

Solution :

Let P(x) = x⁴ + 3x³ + 3x² + x + 1

To find the zero of the ( x + 1), we need to equate it to zero.

x + 1 = 0

x = -1

Given , P(x) = x⁴ + 3x³ + 3x² + x + 1

If (x + 1) is a factor of p (x) = x⁴ + 3x³ + 3x² + x + 1 , p (-1) must be zero

Substituting the value of x in given expression, we get;

P(-1) = (-1)⁴ + 3 ×(-1)³ + 3 ×(-1)² + (-1) + 1

P(-1) = 1 -3 + 3 - 1 + 1

P(-1) = 1

Since, p(–1) ${ \ne }$ 0

Therefore, according to Factor Theorem (x + 1) is not a factor of given polynomial.

Question 1 (iv)

Determine which of the following polynomials has (x + 1) a factor :
(iv) x³ - x²- (2 + √2)x + √2

Solution :

Let P(x) = x³ - x²- (2 + √2)x + √2

To find the zero of the ( x + 1), we need to equate it to zero.

x + 1 = 0

x = -1

Given , P(x) = x³ - x²- (2 + √2)x + √2

If (x + 1) is a factor of p (x) = x³ - x²- (2 + √2)x + √2 , p (-1) must be zero

Substituting the value of x in given expression, we get;

P(-1) = (-1)³ - (-1)²- (2 + √2)(-1) + √2

P(-1) = -1 - 1 + 2 + √2 + √2

P(-1) = 2√2

Since, p(–1) ${ \ne }$ 0

Therefore, according to Factor Theorem (x + 1) is not a factor of given polynomial.

Question 2 (i)

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x³ + x²- 2x -1 , g(x) = x +1

Solution :

Given , P(x) = 2x³ + x²- 2x - 1

To find the zero of the ( x + 1), we need to equate it to zero.

g(x) = 0

x + 1 = 0

x = -1

If (x + 1) is a factor of p (x) = 2x³ + x²- 2x -1 , p (-1) must be zero

Substituting the value of x in given expression, we get;

P(-1) = 2 ×(-1)³ + (-1)²- 2 ×(-1) -1

P(-1) = -2 + 1 + 2 - 1

P(-1) = 0

Since, p(–1) = 0

Therefore, according to Factor Theorem (x + 1) is a factor of given polynomial.

Question 2 (ii)

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(ii) p(x) = x³ + 3x² + 3x + 1 , g(x) = x + 2

Solution :

Given , P(x) = x³ + 3x² + 3x + 1

To find the zero of the ( x + 2), we need to equate it to zero.

g(x) = 0

x + 2 = 0

x = -2

If (x + 2) is a factor of p (x) = x³ + 3x² + 3x + 1 , p (-2) must be zero

Substituting the value of x in given expression, we get;

P(-2) = (-2)³ + 3 ×(-2)² + 3 ×(-2)+ 1

P(-2) = -8 + 12 -6 + 1

P(-2) = -8 + 12 -6 + 1

P(-2) = -1

Since, p(–2) ${ \ne }$ 0

Therefore, according to Factor Theorem (x + 2) is not a factor of given polynomial.

Question 2 (iii)

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(iii) p(x) = x³ - 4x² + x + 6 , g(x) = x - 3

Solution :

Given , P(x) = x³ - 4x² + x + 6

To find the zero of the ( x - 3), we need to equate it to zero.

g(x) = 0

x - 3 = 0

x = 3

If (x - 3) is a factor of p (x) = x³ - 4x² + x + 6 , p (3) must be zero

Substituting the value of x in given expression, we get;

P(3) = (3)³ - 4 × (3)² + 3 + 6

P(3) = 27 - 36 + 3 + 6

P(3) = 36 - 36

P(3) = = 0

Since, p(3) = 0

Therefore, according to Factor Theorem (x - 3) is a factor of given polynomial.

Question 3 (i)

Find the value of k , if x – 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x² + x + k

Solution :

Given , P(x) = x² + x + k

And, ( x – 1 ) is a factor of p(x), then k = ?

To find the zero of the ( x – 1), we need to equate it to zero.

x - 1 = 0

x = 1

If here (x - 1) is a factor of p (x) = x² + x + k , p (1) must be zero

Substituting the value of x in given expression, we get;

P(1) = (1)² + (1) + k

P(1) = 1 + 1 + k

P(1) = 2 + k

Thus, here remainder is , P(1) = 2 + k

Since , x-1 is factor of x² + x + k

According to Factor Theorem , Remainder is zero

So , 2 + k = 0

k = -2

Therefore, K = -2

Question 3 (ii)

Find the value of k , if x – 1 is a factor of p(x) in each of the following cases:
(ii) p(x) = 2x² + kx + √2

Solution :

Given , P(x) = 2x² + kx + √2

And, ( x – 1 ) is a factor of p(x), then k = ?

To find the zero of the ( x – 1), we need to equate it to zero.

x - 1 = 0

x = 1

If here (x - 1) is a factor of p (x) = 2x² + kx + √2 , p (1) must be zero

Substituting the value of x in given expression, we get;

P(1) = 2(1)² + k(1) + √2

P(1) = 2 + k + √2

Thus, here remainder is , P(1) = 2 + k + √2

Since , x-1 is factor of 2x² + kx + √2

According to Factor Theorem , Remainder is zero

So , 2 + k + √2 = 0

k = -2 - √2 or k = -(2 + √2)

Therefore, K = -(2 + √2)

Question 3 (iii)

Find the value of k , if x – 1 is a factor of p(x) in each of the following cases:
(iii) p(x) = kx² -√2x + 1

Solution :

Given , P(x) = kx² -√2x + 1

And, ( x – 1 ) is a factor of p(x), the k = ?

To find the zero of the ( x – 1), we need to equate it to zero.

x - 1 = 0

x = 1

If here (x - 1) is a factor of p (x) = kx² -√2x + 1 , p (1) must be zero

Substituting the value of x in given expression, we get;

P(1) = k(1)² -√2(1) + 1

P(1) = k - √2 + 1

Thus, here remainder is , P(1) = k - √2 + 1

Since , x-1 is factor of kx² -√2x + 1

According to Factor Theorem , Remainder is zero

So , k - √2 + 1 = 0

k = √2 - 1

Therefore, K = √2 - 1

Question 3 (iv)

Find the value of k , if x – 1 is a factor of p(x) in each of the following cases:
(iv) p(x) = kx² -3x + k

Solution :

Given , P(x) = kx² -3x + k

And, ( x – 1 ) is a factor of p(x), the k = ?

To find the zero of the ( x – 1), we need to equate it to zero.

x - 1 = 0

x = 1

If here (x - 1) is a factor of p (x) = kx² -3x + k , p (1) must be zero

Substituting the value of x in given expression, we get;

P(1) = k(1)² -3(1) + k

P(1) = k - 3 + k

P(1) = 2k - 3

Thus, here remainder is , P(1) = 2k - 3

Since , x-1 is factor of kx² -3x + k

According to Factor Theorem , Remainder is zero

So , 2k - 3 = 0

k = 3 / 2

Therefore, K = 3 / 2

Question 4 (i)

Factorise:
(i) 12x² - 7x + 1

Solution :

For quadratic polynomial, We factorize By Splitting the middle term method:

Find 2 numbers p,q such that :

(i) p × q = product of the co-efficient of x² and the constant term (last term)

(ii) p + q = co-efficient of middle term or x

Given , 12x² - 7x + 1

p × q = 12 × 1 = 12

p + q = -7

By Hit and Trial method,

We get p = -3 and q = -4

Now, splitting the middle term of given polynomial can be written as follows:

$$ = 12x^2 - 4x - 3x + 1 $$

( By taking 4x and -1 as common, we get)

$$ = 4x (3x – 1) – 1 (3x – 1) $$

( By taking 3x–1 as common, we get)

$$ = (3x – 1) (4x – 1) $$

Factors of given polynomial : (3x – 1) And (4x – 1)

Question 4 (ii)

Factorise:
(ii) 2x² + 7x + 3

Solution :

For quadratic polynomial, We factorize By Splitting the middle term method:

Find 2 numbers p,q such that :

(i) p × q = product of the co-efficient of x² and the constant term (last term)

(ii) p + q = co-efficient of middle term or x

Given , 2x² + 7x + 3

p × q = 2 × 3 = 6

p + q = 7

By Hit and Trial method,

We get p = 6 and q = 1

Now, splitting the middle term of given polynomial can be written as follows:

$$ = 2x^2 + 6x + x + 3 $$

( By taking 2x and 1 as common, we get)

$$ = 2x (x + 3) + 1 (x + 3) $$

( By taking (x + 3) as common, we get)

$$ = (2x + 1) (x + 3) $$

Factors of given polynomial : (2x + 1) (x + 3)

Question 4 (iii)

Factorise:
(iii) 6x² + 5x - 6

Solution :

For quadratic polynomial, We factorize By Splitting the middle term method:

Find 2 numbers p,q such that :

(i) p × q = product of the co-efficient of x² and the constant term (last term)

(ii) p + q = co-efficient of middle term or x

Given , 6x² + 5x - 6

p × q = 6 × -6 = -36

p + q = 5

By Hit and Trial method,

We get p = 9 and q = -4

Now, splitting the middle term of given polynomial can be written as follows:

$$ = 6x^2 + 9x - 4x - 6 $$

( By taking 3x and -2 as common, we get)

$$ = 3x (2x + 3) - 2 (2x + 3) $$

( By taking (2x + 3) as common, we get)

$$ = (3x - 2) (2x + 3) $$

Factors of given polynomial : (3x - 2) (2x + 3)

Question 4 (iv)

Factorise:
(iv) 3x² - x - 4

Solution :

For quadratic polynomial, We factorize By Splitting the middle term method:

Find 2 numbers p,q such that :

(i) p × q = product of the co-efficient of x² and the constant term (last term)

(ii) p + q = co-efficient of middle term or x

Given , 3x² - x - 4

p × q = 3 × -4 = -12

p + q = -1

By Hit and Trial method,

We get p = 3 and q = -4

Now, splitting the middle term of given polynomial can be written as follows:

$$ = 3x^2 + 3x - 4x - 4 $$

( By taking 3x and -4 as common, we get)

$$ = 3x (x + 1) - 4 (x + 1) $$

( By taking (x + 1) as common, we get)

$$ = (3x - 4) (x + 1) $$

Factors of given polynomial : (3x - 4) (x + 1)

Question 5 (i)

Factorise:
(i) x³ - 2x² - x + 2

Solution :

Given , x³ - 2x² - x + 2

We check integer factors of the constant term, which is 2. The possible factors are = $ \pm $1 and $ \pm $ 2

Now, let p(x) = x³ - 2x² - x + 2

Using Hit and Trial Method, We shall find a factor of p(x) by using some trial value of x, say x =1.

Substituting the value of x = 1 in given expression, we get;

p(1) = (1)³ - (1)x² - (1) + 2

p(1) = 1 - 2 - 1 + 2

p(1) = 0

Hence, p(1) = 0, x =1 is a root of the polynomial.
Therefore, according to Factor Theorem (x – 1) is one of the factor of the given polynomial.

$ \require{enclose} \begin{array}{r|rrrr} & x^2 & -x & -2 \\ \hline x-1 & x^3 & -2x^2 & -x & +2 \\ & x^3 & -x^2 & & \\ \hline & & -x^2 & -x & \\ & & -x^2 & +x & \\ \hline & & & -2x & +2 \\ & & & -2x & +2 \\ \hline & & & & 0 \end{array} $

$$(x - 1 ) ( x^2 - x - 2) $$

Now, factorise the second factor by splitting middle term method

$$ (x^2 - x - 2 ) $$

$$ { x^2 + x - 2x - 2 } $$

$$ {x(x + 1) - 2(x + 1)} $$

$$ {(x - 2)(x + 1) } $$

Factors of given polynomial : (x - 1) (x + 1)(x - 2)

Question 5 (ii)

Factorise:
(ii) x³ - 3x² - 9x - 5

Solution :

Given , x³ - 3x² - 9x - 5

We check integer factors of the constant term, which is -5. The possible factors are = $ \pm 1$ and $ \pm 5$

Now, let p(x) = x³ - 3x² - 9x - 5

Using Hit and Trial Method, We shall find a factor of p(x) by using some trial value of x, say x = 5.

Substituting the value of x = 5 in given expression, we get;

p(5) = (5)³ - 3(5)² - 9(5) - 5

p(5) = 125 - 75 - 45 - 5

p(5) = 0

Hence, p(5) = 0, x = 5 is a root of the polynomial.
Therefore, according to Factor Theorem (x – 5) is one of the factor of the given polynomial.

$ \require{enclose} \begin{array}{r|rrrr} & x^2 & +2x & +1 \\ \hline x-5 & x^3 & -3x^2 & -9x & -5 \\ & x^3 & -5x^2 & & \\ \hline & & 2x^2 & -9x & -5 \\ & & 2x^2 & -10x & \\ \hline & & & x & -5 \\ & & & x & -5 \\ \hline & & & & 0 \end{array} $

$$(x – 5) ( x^2 + 2x + 1) $$

Now, factorise the second factor by splitting middle term method

$$ ( x^2 + 2x + 1) $$

$$ { x^2 + x + x + 1 } $$

$${ x(x + 1) + 1(x + 1)} $$

$$ (x + 1)(x + 1) $$

Factors of given polynomial : (x – 5) (x + 1)(x + 1)

Question 5 (iii)

Factorise:
(iii) x³ + 13x² + 32x + 20

Solution :

Given , x³ + 13x² + 32x + 20

We check integer factors of the constant term, which is 20. The possible factors are = ($ \pm 1$ and $ \pm 20$), ($ \pm 4$ and $ \pm 5$ ), ($ \pm 10$ and $ \pm 2$)

Now, let p(x) = x³ + 13x² + 32x + 20

Using Hit and Trial Method, We shall find a factor of p(x) by using some trial value of x, say x = -1.

Substituting the value of x = -1 in given expression, we get;

p(-1) = (-1)³ + 13(-1)² + 32(-1) + 20

p(-1) = -1 + 13 - 32 + 20

p(-1) = 0

Hence, p(-1) = 0, x = -1 is a root of the polynomial.
Therefore, according to Factor Theorem (x + 1) is one of the factor of the given polynomial.

$ \require{enclose} \begin{array}{r|rrrr} & x^2 & +12x & +20 \\ \hline x+1 & x^3 & +13x^2 & +32x & +20 \\ & x^3 & +x^2 & & \\ \hline & & 12x^2 & +32x & +20 \\ & & 12x^2 & +12x & \\ \hline & & & 20x & +20 \\ & & & 20x & +20 \\ \hline & & & & 0 \end{array} $

$$(x + 1) ( x^2 + 12x + 20) $$

Now, factorise the second factor by splitting middle term method

$$ ( x^2 + 12x + 20) $$

$$ { x^2 + 10x + 2x + 20 } $$

$$ {x(x + 10) + 2(x + 10)} $$

$$ (x + 2)(x + 10) $$

Factors of given polynomial : (x + 1) (x + 2)(x + 10)

Question 5 (iv)

Factorise:
(iv) 2y³ + y² - 2y - 1

Solution :

Given , 2y³ + y² - 2y - 1

We check integer factors of the constant term, which is -1. The possible factors are = $ \pm 1$

Now, let p(y) = 2y³ + y² - 2y - 1

Using Hit and Trial Method, We shall find a factor of p(y) by using some trial value of y, say y = 1.

Substituting the value of y = 1 in given expression, we get;

p(1) = 2(1)³ + (1)² - 2(1) - 1

p(1) = 2 + 1 - 2 -1

p(1) = 0

Hence, p(1) = 0, y = 1 is a root of the polynomial.
Therefore, according to Factor Theorem (y - 1) is one of the factor of the given polynomial.

$ \require{enclose} \begin{array}{r|rrrr} & 2y^2 & +3y & +1 \\ \hline y-1 & 2y^3 & +y^2 & -2y & -1 \\ & 2y^3 & -2y^2 & & \\ \hline & & 3y^2 & -2y & -1 \\ & & 3y^2 & -3y & \\ \hline & & & y & -1 \\ & & & y & -1 \\ \hline & & & & 0 \end{array} $

$$(y - 1) ( 2y^2 + 3y + 1) $$

Now, factorise the second factor by splitting middle term method

$$ ( 2y^2 + 3y + 1) $$

$$ ( 2y^2 + 2y + y + 1) $$

$$ { 2y(y+1)+1(y+1) }$$

$$ (y + 1) (2y + 1) $$

Factors of given polynomial : (y - 1) (y + 1)(2y + 1)