Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.3

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.3

Question 1. Check whether p(x) is a multiple of g(x) or not.
(i) p(x) = \( x^3 – 5x^2 + 4x – 3 \); g(x) = \( x-2 \)
Answer:
p(x) = \( x^3 – 5x^2 + 4x - 3 \)
We need to find p(2) to check if g(x) is a factor. When \( x-2=0 \), then \( x=2 \).
p(2) = \( (2)^3 – 5(2)^2 + 4(2) – 3 \)
= \( 8 – 5(4) + 8 – 3 \)
= \( 8 – 20 + 8 – 3 \)
= \( 16 - 23 \)
= \( -7 \)
Since p(2) \( \neq 0 \), then p(x) is not a multiple of g(x). A polynomial is a multiple of another if the remainder is zero.
In simple words: We check if p(x) can be divided by g(x) without any remainder. If the remainder is not zero (like -7 here), then p(x) is not a multiple of g(x).

🎯 Exam Tip: To check if a polynomial p(x) is a multiple of (x-a), calculate p(a). If p(a) = 0, then (x-a) is a factor and p(x) is a multiple of (x-a).

Question 2. By remainder theorem, find the remainder when p(x) is divided by g(x) where,
(i) p(x) = \( x^3 – 2x^2 – 4x – 1 \); g(x) = \( x + 1 \)
Answer:
p(x) = \( x^3 – 2x^2 – 4x – 1 \)
When p(x) is divided by g(x) = \( x+1 \), we set \( x+1=0 \), which means \( x=-1 \).
Now we find p(-1):
p(-1) = \( (-1)^3 – 2(-1)^2 – 4(-1) – 1 \)
= \( -1 – 2(1) + 4 – 1 \)
= \( -1 – 2 + 4 – 1 \)
= \( -4 + 4 \)
= \( 0 \)
The remainder is 0. This means \( x+1 \) is a factor of p(x).
In simple words: We put \( x=-1 \) into p(x). If the answer is 0, that is the remainder.

🎯 Exam Tip: The Remainder Theorem states that if a polynomial p(x) is divided by \( x-a \), the remainder is p(a). Always remember to change the sign of the constant in g(x) to find the value of x to substitute.

Question 2. By remainder theorem, find the remainder when p(x) is divided by g(x) where,
(ii) p(x) = \( 4x^3 – 12x^2 + 14x – 3 \); g(x) = \( 2x – 1 \)
Answer:
p(x) = \( 4x^3 – 12x^2 + 14x – 3 \)
When p(x) is divided by g(x) = \( 2x-1 \), we set \( 2x-1=0 \), which means \( 2x=1 \), so \( x = \frac{1}{2} \).
Now we find p\( (\frac{1}{2}) \):
p\( (\frac{1}{2}) = 4(\frac{1}{2})^3 – 12(\frac{1}{2})^2 + 14(\frac{1}{2}) – 3 \)
= \( 4(\frac{1}{8}) – 12(\frac{1}{4}) + 14(\frac{1}{2}) – 3 \)
= \( \frac{4}{8} – \frac{12}{4} + \frac{14}{2} – 3 \)
= \( \frac{1}{2} – 3 + 7 – 3 \)
= \( \frac{1}{2} + 1 \)
= \( \frac{3}{2} \)
The remainder is \( \frac{3}{2} \). Fractions often appear as remainders in polynomial division.
In simple words: We put \( x=\frac{1}{2} \) into p(x). The answer we get, \( \frac{3}{2} \), is the remainder.

🎯 Exam Tip: When g(x) is in the form \( ax-b \), remember to substitute \( x = \frac{b}{a} \) into p(x) to find the remainder.

Question 2. By remainder theorem, find the remainder when p(x) is divided by g(x) where,
(iii) p(x) = \( x^3 – 3x^2 + 4x + 50 \); g(x) = \( x – 3 \)
Answer:
p(x) = \( x^3 – 3x^2 + 4x + 50 \)
When p(x) is divided by g(x) = \( x-3 \), we set \( x-3=0 \), which means \( x=3 \).
Now we find p(3):
p(3) = \( (3)^3 – 3(3)^2 + 4(3) + 50 \)
= \( 27 – 3(9) + 12 + 50 \)
= \( 27 – 27 + 12 + 50 \)
= \( 0 + 62 \)
= \( 62 \)
The remainder is 62. This shows that \( x-3 \) is not a factor of p(x).
In simple words: We substitute \( x=3 \) into the polynomial p(x). The number we get, 62, is the remainder when p(x) is divided by \( x-3 \).

🎯 Exam Tip: Always double-check your arithmetic, especially with signs and powers, to avoid errors in calculating p(a).

Question 3. Find the remainder when \( 3x^3 – 4x^2 + 7x – 5 \) is divided by \( (x + 3) \).
Answer:
Let p(x) = \( 3x^3 – 4x^2 + 7x – 5 \).
When p(x) is divided by \( x+3 \), we set \( x+3=0 \), which means \( x=-3 \).
Now we find p(-3):
p(-3) = \( 3(-3)^3 – 4(-3)^2 + 7(-3) – 5 \)
= \( 3(-27) – 4(9) – 21 – 5 \)
= \( -81 – 36 – 21 – 5 \)
= \( -143 \)
The remainder is -143. A negative remainder is common in polynomial division.
In simple words: We put \( x=-3 \) into the polynomial. The final number, -143, is the remainder after division.

🎯 Exam Tip: Be very careful with negative numbers and exponents, as \( (-3)^2 \) is 9, but \( (-3)^3 \) is -27. A common mistake is getting the signs wrong.

Question 4. What is the remainder when \( x^{2018} + 2018 \) is divided by \( x – 1 \).
Answer:
Let p(x) = \( x^{2018} + 2018 \).
When p(x) is divided by \( x-1 \), we set \( x-1=0 \), which means \( x=1 \).
Now we find p(1):
p(1) = \( (1)^{2018} + 2018 \)
= \( 1 + 2018 \)
= \( 2019 \)
The remainder is 2019. Any power of 1 is always 1, simplifying the calculation.
In simple words: We replace \( x \) with 1 in the polynomial. Since 1 raised to any power is still 1, the remainder is 1 plus 2018, which is 2019.

🎯 Exam Tip: For expressions involving very large exponents, substituting x=1 or x=-1 often simplifies the calculation significantly, as \( 1^n=1 \) and \( (-1)^n \) is either 1 or -1.

Question 5. For what value of k is the polynomial p(x) = \( 2x^3 – kx^2 + 3x + 10 \) exactly divisible by \( x – 2 \).
Answer:
Let p(x) = \( 2x^3 – kx^2 + 3x + 10 \).
For p(x) to be exactly divisible by \( x-2 \), the remainder must be 0. So, p(2) = 0.
p(2) = \( 2(2)^3 – k(2)^2 + 3(2) + 10 = 0 \)
\( 2(8) – k(4) + 6 + 10 = 0 \)
\( 16 – 4k + 6 + 10 = 0 \)
\( 32 – 4k = 0 \)
Now, we solve for k:
\( 32 = 4k \)
\( k = \frac{32}{4} \)
\( k = 8 \)
The value of k is 8. When k is 8, the polynomial can be divided by \( x-2 \) without any remainder.
In simple words: For the polynomial to divide perfectly by \( x-2 \), we must get a remainder of zero when we put \( x=2 \) into it. We use this rule to find the missing number, k.

🎯 Exam Tip: When a polynomial is "exactly divisible" by a factor, it means the remainder is 0. Set p(a) = 0 and solve for the unknown variable.

Question 6. If two polynomials \( 2x^3 + ax^2 + 4x – 12 \) and \( x^3 + x^2 - 2x + a \) leave the same remainder when divided by \( (x – 3) \), find the value of a and also find the remainder.
Answer:
Let p\( _1 \)(x) = \( 2x^3 + ax^2 + 4x – 12 \).
When p\( _1 \)(x) is divided by \( x-3 \), the remainder is p\( _1 \)(3).
p\( _1 \)(3) = \( 2(3)^3 + a(3)^2 + 4(3) – 12 \)
= \( 2(27) + 9a + 12 – 12 \)
= \( 54 + 9a \) ..........(R\( _1 \))

Let p\( _2 \)(x) = \( x^3 + x^2 - 2x + a \).
When p\( _2 \)(x) is divided by \( x-3 \), the remainder is p\( _2 \)(3).
p\( _2 \)(3) = \( (3)^3 + (3)^2 – 2(3) + a \)
= \( 27 + 9 – 6 + a \)
= \( 36 – 6 + a \)
= \( 30 + a \) ..........(R\( _2 \))

The problem states that the remainders are the same (R\( _1 \) = R\( _2 \)):
\( 54 + 9a = 30 + a \)
Now, we solve for a:
\( 9a – a = 30 – 54 \)
\( 8a = -24 \)
\( a = \frac{-24}{8} \)
\( a = -3 \)
The value of a is -3. This makes the remainders equal for both polynomials.

To find the remainder, substitute \( a=-3 \) into R\( _2 \) (or R\( _1 \)):
Remainder = \( 30 + a \)
= \( 30 + (-3) \)
= \( 27 \)
The remainder is 27.
In simple words: We calculate the remainder for each polynomial when divided by \( x-3 \). We get two expressions, one for each remainder. Since the problem says these remainders are equal, we set the two expressions equal to each other to find 'a'. Then we use the value of 'a' to find the actual remainder.

🎯 Exam Tip: Organize your work by clearly labeling each polynomial and its remainder expression. This helps avoid confusion when dealing with multiple functions and variables.

Question 7. Determine whether \( (x – 1) \) is a factor of the following polynomials:
(i) p(x) = \( x^3 + 5x^2 – 10x + 4 \)
Answer:
For \( (x-1) \) to be a factor, p(1) must be 0.
p(1) = \( (1)^3 + 5(1)^2 – 10(1) + 4 \)
= \( 1 + 5(1) – 10 + 4 \)
= \( 1 + 5 – 10 + 4 \)
= \( 10 – 10 \)
= \( 0 \)
Since p(1) = 0, \( (x-1) \) is a factor of p(x). When the remainder is zero, it means the division is exact.
In simple words: We put \( x=1 \) into the polynomial. If the answer is 0, then \( x-1 \) is a factor.

🎯 Exam Tip: The Factor Theorem is a special case of the Remainder Theorem: if p(a)=0, then \( x-a \) is a factor of p(x).

Question 7. Determine whether \( (x – 1) \) is a factor of the following polynomials:
(ii) p(x) = \( x^4 + 5x^2 – 5x + 1 \)
Answer:
For \( (x-1) \) to be a factor, p(1) must be 0.
p(1) = \( (1)^4 + 5(1)^2 – 5(1) + 1 \)
= \( 1 + 5(1) – 5 + 1 \)
= \( 1 + 5 – 5 + 1 \)
= \( 7 – 5 \)
= \( 2 \)
Since p(1) \( \neq 0 \), \( (x-1) \) is not a factor of p(x). The remainder is 2.
In simple words: We put \( x=1 \) into the polynomial. Since the answer is 2 (not 0), then \( x-1 \) is not a factor.

🎯 Exam Tip: Clearly state whether \( x-a \) is a factor or not based on whether p(a) is 0. Don't just show the calculation without a conclusion.

Question 8. Using factor theorem, show that \( (x – 5) \) is a factor of the polynomial \( 2x^3 - 5x^2 – 28x + 15 \).
Answer:
Let p(x) = \( 2x^3 - 5x^2 – 28x + 15 \).
According to the Factor Theorem, if \( (x-5) \) is a factor, then p(5) must be 0.
p(5) = \( 2(5)^3 – 5(5)^2 – 28(5) + 15 \)
= \( 2(125) – 5(25) – 140 + 15 \)
= \( 250 – 125 – 140 + 15 \)
= \( 265 – 265 \)
= \( 0 \)
Since p(5) = 0, \( (x-5) \) is a factor of p(x). This confirms the statement in the question.
In simple words: We need to prove that \( x-5 \) divides the polynomial exactly. We do this by putting \( x=5 \) into the polynomial. If the final answer is zero, then \( x-5 \) is indeed a factor.

🎯 Exam Tip: "Show that" questions require you to present the calculations clearly and explicitly state the conclusion based on the result (p(a)=0 or p(a)≠0).

Question 9. Determine the value of m, if \( (x + 3) \) is a factor of \( x^3 – 3x^2 – mx + 24 \).
Answer:
Let p(x) = \( x^3 – 3x^2 – mx + 24 \).
Since \( (x+3) \) is a factor, according to the Factor Theorem, p(-3) must be 0.
p(-3) = \( (-3)^3 – 3(-3)^2 – m(-3) + 24 = 0 \)
\( -27 – 3(9) + 3m + 24 = 0 \)
\( -27 – 27 + 3m + 24 = 0 \)
\( -54 + 3m + 24 = 0 \)
\( -30 + 3m = 0 \)
Now, we solve for m:
\( 3m = 30 \)
\( m = \frac{30}{3} \)
\( m = 10 \)
The value of m is 10. This ensures that \( x+3 \) is a factor of the polynomial.
In simple words: Because \( x+3 \) is a factor, putting \( x=-3 \) into the polynomial must give an answer of zero. We use this fact to create an equation and find the value of the unknown 'm'.

🎯 Exam Tip: When given that a term is a factor, always remember to set the polynomial equal to zero after substitution to form an equation for the unknown variable.

Question 10. If both \( (x-2) \) and \( (x - \frac{1}{2}) \) are the factors of \( ax^2 + 5x + b \), then show that a = b.
Answer:
Let p(x) = \( ax^2 + 5x + b \).
Since \( (x-2) \) is a factor, p(2) must be 0:
p(2) = \( a(2)^2 + 5(2) + b = 0 \)
\( 4a + 10 + b = 0 \)
\( 4a + b = -10 \) .......(1)

Since \( (x - \frac{1}{2}) \) is also a factor, p\( (\frac{1}{2}) \) must be 0:
p\( (\frac{1}{2}) = a(\frac{1}{2})^2 + 5(\frac{1}{2}) + b = 0 \)
\( a(\frac{1}{4}) + \frac{5}{2} + b = 0 \)
To remove fractions, multiply the entire equation by 4:
\( 4 \times (\frac{a}{4}) + 4 \times (\frac{5}{2}) + 4 \times b = 4 \times 0 \)
\( a + 10 + 4b = 0 \)
\( a + 4b = -10 \) .......(2)

Now, we have two equations:
1) \( 4a + b = -10 \)
2) \( a + 4b = -10 \)

Since both equations equal -10, we can set them equal to each other:
\( 4a + b = a + 4b \)
Now, we rearrange to show a = b:
\( 4a – a = 4b – b \)
\( 3a = 3b \)
Divide both sides by 3:
\( a = b \)
Hence, it is proved that a = b. This result holds true for any polynomial that satisfies the given conditions.
In simple words: We use the factor theorem for both given factors. This gives us two equations, both equal to -10. We set these two equations equal to each other and simplify them to show that 'a' must be the same as 'b'.

🎯 Exam Tip: When dealing with multiple factors, create a separate equation for each factor. Then use simultaneous equations to solve for unknowns or prove relationships like a=b.

Question 11. If \( (x – 1) \) divides the polynomial \( kx^3 – 2x^2 + 25x – 26 \) without remainder, then find the value of k.
Answer:
Let p(x) = \( kx^3 – 2x^2 + 25x – 26 \).
If \( (x-1) \) divides p(x) without remainder, it means \( x-1 \) is a factor, so p(1) must be 0.
p(1) = \( k(1)^3 – 2(1)^2 + 25(1) – 26 = 0 \)
\( k(1) – 2(1) + 25 – 26 = 0 \)
\( k – 2 + 25 – 26 = 0 \)
\( k + 25 – 28 = 0 \)
\( k – 3 = 0 \)
Now, we solve for k:
\( k = 3 \)
The value of k is 3. This is the only value for which \( x-1 \) will divide the polynomial exactly.
In simple words: Since \( x-1 \) divides the polynomial perfectly, putting \( x=1 \) into the polynomial must give a remainder of zero. We use this to solve for the unknown 'k'.

🎯 Exam Tip: Pay close attention to the wording "without remainder" as it directly implies that the Factor Theorem should be applied, setting the substituted polynomial value to zero.

Question 12. Check if \( (x + 2) \) and \( (x – 4) \) are the sides of a rectangle whose area is \( x^2 – 2x – 8 \) by using factor theorem.
Answer:
Let the area of the rectangle be p(x) = \( x^2 – 2x – 8 \).
If \( (x+2) \) is a side (factor), then p(-2) must be 0.
p(-2) = \( (-2)^2 – 2(-2) – 8 \)
= \( 4 + 4 – 8 \)
= \( 8 – 8 \)
= \( 0 \)
Since p(-2) = 0, \( (x+2) \) is a factor of p(x).

If \( (x-4) \) is a side (factor), then p(4) must be 0.
p(4) = \( (4)^2 – 2(4) – 8 \)
= \( 16 – 8 – 8 \)
= \( 16 – 16 \)
= \( 0 \)
Since p(4) = 0, \( (x-4) \) is also a factor of p(x).
Since both \( (x+2) \) and \( (x-4) \) are factors of the area polynomial, they can represent the sides of the rectangle. The area of a rectangle is length times width, which means the sides are factors of the area expression.
In simple words: The area of a rectangle is found by multiplying its sides. So, the sides must be factors of the area polynomial. We use the factor theorem by checking if putting \( x=-2 \) and \( x=4 \) into the area polynomial gives zero. Since both give zero, both \( (x+2) \) and \( (x-4) \) are factors, meaning they can be the sides.

🎯 Exam Tip: For geometric problems involving polynomial factors, relate the geometric property (like area being product of sides) to the algebraic concept of factors and zeros of a polynomial.