GSEB Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.4

Question 1. Determine which of the following polynomials has \( x + 1 \) a factor:
1. \( x^3 + x^2 + x + 1 \)
2. \( x^4 + x^3 + x^2 + x + 1 \)
3. \( x^4 + 3x^3 + 3x^2 + x + 1 \)
4. \( x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2} \)
Answer:
1. Let \( p(x) = x^3 + x^2 + x + 1 \).
Set \( x + 1 = 0 \)
So, \( x = -1 \). (If \( x + 1 \) is a factor of \( p(x) \), then \( x + 1 \) will be equal to zero)
Now, substitute \( x = -1 \) into \( p(x) \):
\( p(-1) = (-1)^3 + (-1)^2 + (-1) + 1 \)
\( = -1 + 1 - 1 + 1 \)
\( = 0 \)
Thus, by the factor theorem, \( x + 1 \) is a factor of \( x^3 + x^2 + x + 1 \).
2. Let \( p(x) = x^4 + x^3 + x^2 + x + 1 \).
Set \( x + 1 = 0 \)
So, \( x = -1 \).
Now, substitute \( x = -1 \) into \( p(x) \):
\( p(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 \)
\( = 1 - 1 + 1 - 1 + 1 \)
\( = 1 \)
Since \( 1 \neq 0 \), by the factor theorem, \( x + 1 \) is not a factor of \( x^4 + x^3 + x^2 + x + 1 \).
3. Let \( p(x) = x^4 + 3x^3 + 3x^2 + x + 1 \).
Set \( x + 1 = 0 \)
So, \( x = -1 \).
Now, substitute \( x = -1 \) into \( p(x) \):
\( p(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1 \)
\( = 1 + 3(-1) + 3(1) - 1 + 1 \)
\( = 1 - 3 + 3 - 1 + 1 \)
\( = 1 \)
Since \( 1 \neq 0 \), by the factor theorem, \( x + 1 \) is not a factor of \( x^4 + 3x^3 + 3x^2 + x + 1 \).
4. Let \( p(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2} \).
Set \( x + 1 = 0 \)
So, \( x = -1 \).
Now, substitute \( x = -1 \) into \( p(x) \):
\( p(-1) = (-1)^3 - (-1)^2 - (2 + \sqrt{2})(-1) + \sqrt{2} \)
\( = -1 - 1 + (2 + \sqrt{2}) + \sqrt{2} \)
\( = -2 + 2 + \sqrt{2} + \sqrt{2} \)
\( = 2\sqrt{2} \)
Since \( 2\sqrt{2} \neq 0 \), by the factor theorem, \( x + 1 \) is not a factor of \( x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2} \).
In simple words: To find out if \( x + 1 \) is a factor, we set \( x + 1 \) to zero to get \( x = -1 \). Then, we put this value into each polynomial. If the answer is zero, \( x + 1 \) is a factor; if not, it isn't.

Exam Tip: Remember the Factor Theorem states that \( (x-a) \) is a factor of a polynomial \( p(x) \) if and only if \( p(a) = 0 \). Pay careful attention to signs when substituting negative values.

Question 2. Use the factor theorem to determine whether \( g(x) \) is a factor of \( p(x) \) in each of the following cases:
1. \( p(x) = 2x^3 + x^2 - 2x - 1 \), \( g(x) = x + 1 \)
2. \( p(x) = x^3 + 3x^2 + 3x + 1 \), \( g(x) = x + 2 \)
3. \( p(x) = x^3 - 4x^2 + x + 6 \), \( g(x) = x - 3 \)
Answer:
1. Let \( p(x) = 2x^3 + x^2 - 2x - 1 \)
Set \( g(x) = 0 \)
\( x + 1 = 0 \)
So, \( x = -1 \).
Now, substitute \( x = -1 \) into \( p(x) \):
\( p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1 \)
\( = 2(-1) + (1) - 2(-1) - 1 \)
\( = -2 + 1 + 2 - 1 \)
\( = 0 \)
Thus, by the factor theorem, \( g(x) \) is a factor of \( p(x) \).
2. Let \( p(x) = x^3 + 3x^2 + 3x + 1 \)
Set \( g(x) = 0 \)
\( x + 2 = 0 \)
So, \( x = -2 \).
Now, substitute \( x = -2 \) into \( p(x) \):
\( p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1 \)
\( = -8 + 3(4) - 6 + 1 \)
\( = -8 + 12 - 6 + 1 \)
\( = -1 \)
Since \( -1 \neq 0 \), by the factor theorem, \( g(x) \) is not a factor of \( p(x) \).
3. Let \( p(x) = x^3 - 4x^2 + x + 6 \)
Set \( g(x) = 0 \)
\( x - 3 = 0 \)
So, \( x = 3 \).
Now, substitute \( x = 3 \) into \( p(x) \):
\( p(3) = (3)^3 - 4(3)^2 + 3 + 6 \)
\( = 27 - 4(9) + 9 \)
\( = 27 - 36 + 9 \)
\( = 0 \)
Thus, by the factor theorem, \( g(x) \) is a factor of \( p(x) \).
In simple words: For each part, we find the value of \( x \) that makes \( g(x) \) zero. Then we put that \( x \) value into \( p(x) \). If the final result is zero, then \( g(x) \) divides \( p(x) \) perfectly, meaning it's a factor.

Exam Tip: Carefully follow the steps of the Factor Theorem by finding the root of \( g(x) \) and substituting it into \( p(x) \). Be very careful with integer arithmetic, especially when dealing with negative numbers raised to powers.

Question 3. Find the value of k, if \( x - 1 \) is a factor of \( p(x) \) in each of the following cases:
1. \( p(x) = x^2 + x + k \)
2. \( p(x) = 2x^2 + kx + \sqrt{2} \)
3. \( p(x) = kx^2 - \sqrt{2}x + 1 \)
4. \( p(x) = kx^2 - 3x + k \)
Answer:
Since \( x - 1 \) is a factor of \( p(x) \), by the factor theorem, \( p(1) = 0 \).
1. Let \( p(x) = x^2 + x + k \).
Substitute \( x = 1 \):
\( p(1) = (1)^2 + 1 + k \)
\( 0 = 1 + 1 + k \)
\( 0 = 2 + k \)
Therefore, \( k = -2 \).
2. Let \( p(x) = 2x^2 + kx + \sqrt{2} \).
Substitute \( x = 1 \):
\( p(1) = 2(1)^2 + k(1) + \sqrt{2} \)
\( 0 = 2(1) + k + \sqrt{2} \)
\( 0 = 2 + k + \sqrt{2} \)
Therefore, \( k = -2 - \sqrt{2} \).
\( k = -(2 + \sqrt{2}) \)
3. Let \( p(x) = kx^2 - \sqrt{2}x + 1 \).
Substitute \( x = 1 \):
\( p(1) = k(1)^2 - \sqrt{2}(1) + 1 \)
\( 0 = k - \sqrt{2} + 1 \)
Therefore, \( k = \sqrt{2} - 1 \).
4. Let \( p(x) = kx^2 - 3x + k \).
Substitute \( x = 1 \):
\( p(1) = k(1)^2 - 3(1) + k \)
\( 0 = k - 3 + k \)
\( 0 = 2k - 3 \)
\( 2k = 3 \)
Therefore, \( k = \frac{3}{2} \).
In simple words: If \( x - 1 \) is a factor, it means that when you put \( x = 1 \) into the polynomial, the answer must be zero. We use this fact to solve for the unknown number \( k \) in each polynomial.

Exam Tip: This type of problem directly tests your understanding of the Factor Theorem. Setting \( p(1)=0 \) is the key first step. Always double-check your algebraic manipulations when solving for \( k \).

Question 4. Factorise:
1. \( 12x^2 - 7x + 1 \)
2. \( 2x^2 + 7x + 3 \)
3. \( 6x^2 + 5x - 6 \)
4. \( 3x^2 - x - 4 \)
Answer:
1. To factorise \( 12x^2 - 7x + 1 \):
We look for two numbers that multiply to \( (a \times c) = 12 \times 1 = 12 \) and add to \( b = -7 \). These numbers are \( -4 \) and \( -3 \).
We rewrite the middle term \( -7x \) as \( -4x - 3x \).
\( = 12x^2 - 4x - 3x + 1 \)
Now, group the terms and factor out common parts:
\( = 4x(3x - 1) - 1(3x - 1) \)
\( = (3x - 1)(4x - 1) \)
2. To factorise \( 2x^2 + 7x + 3 \):
We look for two numbers that multiply to \( (a \times c) = 2 \times 3 = 6 \) and add to \( b = 7 \). These numbers are \( 6 \) and \( 1 \).
We rewrite the middle term \( 7x \) as \( 6x + x \).
\( = 2x^2 + 6x + x + 3 \)
Now, group the terms and factor out common parts:
\( = 2x(x + 3) + 1(x + 3) \)
\( = (x + 3)(2x + 1) \)
3. To factorise \( 6x^2 + 5x - 6 \):
We look for two numbers that multiply to \( (a \times c) = 6 \times (-6) = -36 \) and add to \( b = 5 \). These numbers are \( 9 \) and \( -4 \).
We rewrite the middle term \( 5x \) as \( 9x - 4x \).
\( = 6x^2 + 9x - 4x - 6 \)
Now, group the terms and factor out common parts:
\( = 3x(2x + 3) - 2(2x + 3) \)
\( = (2x + 3)(3x - 2) \)
4. To factorise \( 3x^2 - x - 4 \):
We look for two numbers that multiply to \( (a \times c) = 3 \times (-4) = -12 \) and add to \( b = -1 \). These numbers are \( -4 \) and \( 3 \).
We rewrite the middle term \( -x \) as \( -4x + 3x \).<{"type": "text", "text_content": "\nNow, group the terms and factor out common parts:\n\( = x(3x - 4) + 1(3x - 4) \)\n\( = (3x - 4)(x + 1) \)
In simple words: To factorise these expressions, we split the middle term into two parts. These two parts should add up to the middle term and multiply to the product of the first and last terms. Then, we group the terms and find the common factors.

Exam Tip: When factorising quadratic expressions like \( ax^2 + bx + c \), always first look for a common factor. If none, use the splitting the middle term method. The two numbers must multiply to \( ac \) and add to \( b \).

Question 5. Factorise:
1. \( x^3 - 2x^2 - x + 2 \)
2. \( x^3 - 3x^2 - 9x - 5 \)
3. \( x^3 + 13x^2 + 32x + 20 \)
4. \( 2y^3 + y^2 - 2y - 1 \)
Answer:
1. To factorise \( p(x) = x^3 - 2x^2 - x + 2 \):
The constant term is 2, and its factors are \( \pm 1, \pm 2 \). We use the trial method to find a factor.
Let's test \( x = 1 \):
\( p(1) = (1)^3 - 2(1)^2 - 1 + 2 \)
\( = 1 - 2 - 1 + 2 \)
\( = 3 - 3 \)
\( = 0 \)
Since \( p(1) = 0 \), \( (x - 1) \) is a factor of \( p(x) \).
Now, we divide \( p(x) \) by \( (x - 1) \) or use grouping:
\( x^3 - 2x^2 - x + 2 \)
\( = x^2(x - 1) - x(x - 1) - 2(x - 1) \) (This step seems incorrect, let's follow the OCR grouping shown after the error)
\( = x^2 - x^2 + x - 2x + 2 \) (This line is not logical. Let's assume the subsequent steps are for a correct grouping strategy to follow the problem correctly. OCR seems to have made an error with the previous line)
\( = x(x - 1) - x(x - 1) - 2(x - 1) \) (This also seems incorrect. Let's use the valid sequence of steps from the OCR.)
The OCR solution appears to contain an error in the grouping step for the first sub-part, but then provides a correct factored form based on other methods. I will reproduce the final factored form by grouping from the correct factorization steps in the OCR for consistency with the intended solution. The solution shows:
\( x^3 - 2x^2 - x + 2 \)
\( = x^2(x - 2) - 1(x - 2) \)
\( = (x - 2)(x^2 - 1) \)
\( = (x - 2)(x - 1)(x + 1) \)
2. To factorise \( p(x) = x^3 - 3x^2 - 9x - 5 \):
The constant term is \( -5 \), and its factors are \( \pm 1, \pm 5 \). We use the trial method.
Let's test \( x = 1 \):
\( p(1) = (1)^3 - 3(1)^2 - 9(1) - 5 \)
\( = 1 - 3 - 9 - 5 \)
\( = -16 \)
Since \( p(1) \neq 0 \), \( (x - 1) \) is not a factor.
Let's test \( x = -1 \):
\( p(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 \)
\( = -1 - 3(1) - 9(-1) - 5 \)
\( = -1 - 3 + 9 - 5 \)
\( = 0 \)
Since \( p(-1) = 0 \), \( (x + 1) \) is a factor of \( p(x) \).
Now, we group the terms:
\( x^3 - 3x^2 - 9x - 5 \)
\( = x^3 + x^2 - 4x^2 - 4x - 5x - 5 \)
\( = x^2(x + 1) - 4x(x + 1) - 5(x + 1) \)
\( = (x + 1)(x^2 - 4x - 5) \)
Now, factorise the quadratic term \( x^2 - 4x - 5 \). We look for two numbers that multiply to \( -5 \) and add to \( -4 \). These are \( -5 \) and \( 1 \).
\( = (x + 1)[x^2 - 5x + x - 5] \)
\( = (x + 1)[x(x - 5) + 1(x - 5)] \)
\( = (x + 1)(x - 5)(x + 1) \)
\( = (x + 1)^2(x - 5) \)
3. To factorise \( p(x) = x^3 + 13x^2 + 32x + 20 \):
The constant term is 20, and its factors can be \( \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20 \). We use the trial method.
Let's test \( x = -1 \):
\( p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20 \)
\( = -1 + 13(1) - 32 + 20 \)
\( = -1 + 13 - 32 + 20 \)
\( = 33 - 33 \)
\( = 0 \)
Since \( p(-1) = 0 \), \( (x + 1) \) is a factor of \( p(x) \).
Now, we group the terms:
\( x^3 + 13x^2 + 32x + 20 \)
\( = x^3 + x^2 + 12x^2 + 12x + 20x + 20 \)
\( = x^2(x + 1) + 12x(x + 1) + 20(x + 1) \)
\( = (x + 1)(x^2 + 12x + 20) \)
Now, factorise the quadratic term \( x^2 + 12x + 20 \). We look for two numbers that multiply to 20 and add to 12. These are \( 10 \) and \( 2 \).
\( = (x + 1)[x^2 + 10x + 2x + 20] \)
\( = (x + 1)[x(x + 10) + 2(x + 10)] \)
\( = (x + 1)(x + 10)(x + 2) \)
4. To factorise \( p(y) = 2y^3 + y^2 - 2y - 1 \):
The constant term is \( -1 \), and its factors are \( \pm 1 \). We use the trial method.
Let's test \( y = 1 \):
\( p(1) = 2(1)^3 + (1)^2 - 2(1) - 1 \)
\( = 2 + 1 - 2 - 1 \)
\( = 3 - 3 \)
\( = 0 \)
Since \( p(1) = 0 \), \( (y - 1) \) is a factor of \( p(y) \).
Now, we group the terms:
\( 2y^3 + y^2 - 2y - 1 \)
\( = 2y^3 - 2y^2 + 3y^2 - 3y + y - 1 \)
\( = 2y^2(y - 1) + 3y(y - 1) + 1(y - 1) \)
\( = (y - 1)(2y^2 + 3y + 1) \)
Now, factorise the quadratic term \( 2y^2 + 3y + 1 \). We look for two numbers that multiply to \( 2 \times 1 = 2 \) and add to 3. These are \( 2 \) and \( 1 \).
\( = (y - 1)[2y^2 + 2y + y + 1] \)
\( = (y - 1)[2y(y + 1) + 1(y + 1)] \)
\( = (y - 1)(y + 1)(2y + 1) \)
In simple words: For these cubic polynomials, we first use trial and error with factors of the constant term to find one root. This gives us one factor. Then, we can use grouping or polynomial division to find the other factors and fully break down the polynomial into its simpler parts.

Exam Tip: For factorising cubic polynomials, the Factor Theorem is crucial. Once a factor is found, polynomial division or strategic grouping of terms helps to reduce the polynomial to a quadratic, which can then be factorised further. Practice recognising common factor patterns.