5 Marks Questions

Question 15 Marks

If p(y) = 4 - 3y -y² + 5y³, find:

p(0)
p(2)
p(-1)

Answer

p(y) = 4 - 3y -y² + 5y³

⇒ p(0) = (4 + 3 × 0 - 0² + 5 × 0³ )

= (4 + 0 - 0 + 0)

= 4

p(y) = 4 - 3y -y² + 5y³

⇒ p(2) = (4 + 3 × 2 - 2² + 5 × 2³ )

= (4 + 6 - 4 + 40)

= 46

p(y) = 4 - 3y -y² + 5y³

⇒ p(-1) = [(4 + 3 ×(- 1)² + 5 × (-1)³ )]

= (4 - 3 - 1 - 5)

= -5

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Question 25 Marks

If p(x) = 5 - 4x + 2x², find:

p(0)
p(3)
p(-2)

Answer

p(x) = 5 - 4x + 2x²

⇒ p(0) = (5 - 4 × 0 + 2 × 0²)

= (5 - 0 + 0)

= 5

p(x) = 5 - 4x + 2x²

p(3) = (5 - 4 × 3 + 2 × 3²)

= (5 - 12 + 18)

= 11

p(x) = 5 - 4x + 2x²

⇒ p(-2) = [(5 - 4 × (-2) + 2 × (-2)²]

= (5 + 8 + 8)

= 21

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Question 35 Marks

If f(t) = 4t² - 3t + 6, find:

f(0)
f(4)
f(-5)

Answer

f(t) = 4t² - 3t + 6

⇒ f(0) = (4 × 0² - 3 × 0 + 6)

= (0 - 0 + 6)

= 6

f(t) = 4t² - 3t + 6

⇒ f(4) = (4 × 4² - 3 × 4 + 6)

= (64 - 12 + 6)

= 58

f(t) = 4t² - 3t + 6

⇒ f(-5) = [(4 × (-5)² - 3 × (-5) + 6)]

= (100 + 15 + 6)

= 121

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Question 45 Marks

If both (x - 2) and $\Big(\text{x}-\frac{1}{2}\Big)$ are factors of px² + 5x + r, prove that p = r.

Answer

Let f(x) = px² + 5x + r
Now, (x - 2) is a factor of f(x).
⇒ f(2) = 0
⇒ p(2)² + 5(2) + r = 0
⇒ 4p + 10 + r = 0
⇒ 4p + r = -10 ...(i)
Also, $\Big(\text{x}-\frac{1}{2}\Big)$ is a factor of f(x).
$\Rightarrow\text{f}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow\text{p}\Big(\frac{1}{2}\Big)+5\times\frac{1}{2}+\text{r}=0$
$\Rightarrow\frac{\text{p}}{4}+\frac{5}{2}+\text{r}=0$
$\Rightarrow\frac{\text{p}+10+4\text{r}}{4}=0$
$\Rightarrow\text{p}+4\text{r}+10=0$
$\Rightarrow\text{p}+4\text{r}=-10\ ...(\text{ii})$
From (i) and (ii), we have
4p + r = p + 4r
⇒ 4p - p = 4r - r
⇒ 3p = 3r
⇒ p = r

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Question 55 Marks

If p(x) = x³ - 3x² + 2x, find p(0), p(1), p(2). What do you conclude?

Answer

p(x) = x³ - 3x² + 2x ...(1)
Putting x = 0 in (1), we get
p(0) = 0³ - 3 × 0² + 2 × 0 = 0
Thus, x = 0 is a zero of p(x).
Putting x = 1 in (1), we get
p(1) = 1³ - 3 × 1² + 2 × 1
= 1 - 3 + 2 = 0
Thus, x = 1 is a zero of p(x).
Putting x = 2 in (1), we get
p(2) = 2³ - 3 × 2² + 2 × 2
= 8 - 3 × 4 + 4
= 8 - 12 + 4 = 0
Thus, x = 2 is a zero of p(x).

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Question 65 Marks

Verify the division algorithm for the polynomials
p(x) = 2x⁴ - 6x³ + 2x² - x + 2 and g(x) = x + 2.

Answer

p(x) = 2x⁴ - 6x³ + 2x² - x + 2 and g(x) = x + 2

Quotient = 2x³ - 10x² + 22x - 45

Remainder = 92

Verification:

Divisor × Quotient + Remainder

= (x + 2) × (2x³ - 10x² + 22x - 45) + 92

= x (2x³ - 10x² + 22x - 45) + 2(2x³ - 10x² + 22x - 45) + 92

= 2x⁴ - 10x³ + 22x² - 45x + 4x³ - 20x² + 44x - 90 + 92

= 2x⁴ - 6x³ + 2x² - x + 2

= Dividend

Hence verified.

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Question 75 Marks

Find the values of a and b so that the polynomial (x⁴ + ax³ - 7x²- 8x + b) is exactly divisible by (x + 2) as well as (x + 3).

Answer

Let f(x)= (x⁴ + ax³ - 7x² - 8x + b)
Now, x + 2 = 0
⇒ x = -2 and,
⇒ x + 3 = 0
⇒ x = -3
By factor theorem, (x + 2) and (x + 3) will be factors of f(x) if f(-2) = 0 and f(-3) = 0
$\therefore$ f(-2) = (-2)⁴ + a(-2)³ - 7(-2)² - 8(-2) + b = 0
⇒ 16 - 8a - 28 + 16 + b = 0
⇒ -8a + b = -4
⇒ 8a - b = 4 ...(i)
And, f(-3) = (-3)⁴ + a(-3)³ - 7(-3)² - 8(-3) + b = 0
⇒ 81 - 27a - 63 + 24 + b = 0
⇒ -27a + b = -42
⇒ 27a - b = 42 ...(ii)
Subtracting (i) from (ii), we get,
19a = 38
So, a = 2
Substituting the value of a = 2 in (i), we get
8 × 2 - b = 4
⇒ 16 - b = 4
⇒ -b = -16 + 4
⇒ -b = -12
⇒ b = 12
$\therefore$ a = 2 and b = 12.

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Question 85 Marks

If 2 and 0 are the zeros of the polynomial f(x) = 2x³ - 5x² + ax + b then find the values of a and b.
Hint: f(x) = 0 and f(0) = 0.

Answer

It is given that 2 and 0 are the zeros of the polynomial f(x) = 2x³ - 5x² + ax + b.
$\therefore$ f(2) = 0
⇒ 2 × 2³ - 5 × 2² + a × 2 + b = 0
⇒ 16 - 20 + 2a + b = 0
⇒ -4 + 2a + b = 0
⇒ 2a + b = 4 ...(1)
Also,
f(0) = 0
⇒ 2 × 0³ - 5 × 0² + a × 0 + b = 0
⇒ 0 - 0 + 0 + b = 0
⇒ b = 0
Putting b = 0 in (1), we get
2a + 0 = 4
⇒ 2a = 4
⇒ a = 2
Thus, the values of a and b are 2 and 0, respectively.

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Question 95 Marks

If p(x) = x³ + x² - 9x - 9, find p(0), p(3), p(-3) and p(-1). What do you conclude about the zeros of p(x)? Is 0 a zero of p(x)?

Answer

p(x) = x³ + x² - 9x - 9 ...(1)
Putting x = 0 in (1), we get
p(0) = 0³ + 0² - 9 × 0 - 9
$=-9\neq0$
Thus, x = 0 is not a zero of p(x).
Putting x = 3 in (1), we get
p(3) = 3³ + 3² - 9 × 3 - 9
= 27 + 9 - 27 - 9 = 0
Thus, x = 3 is a zero of p(x).
Putting x = -3 in (1), we get
p(-3) = (-3)³ + (-3)² - 9 × (-3) - 9
= -27 + 9 + 27 - 9 = 0
Thus, x = -3 is a zero of p(x).
Putting x = -1 in (1), we get
p(-1) = (-1)³ + (-1)² - 9 × (-1) - 9
= -1 + 1 + 9 - 9 = 0
Thus, x = -1 is a zero of p(x).

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Question 105 Marks

The polynomial p(x) = x⁴ - 2x³ + 3x² - ax + b when divided by (x - 1) and (x + 1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when p(x) is divided by (x - 2).

Answer

Let:
p(x) = x⁴ - 2x³ + 3x² - ax + b
Now,
When p(x) is divided by (x - 1), the remainder is p(1).
When p(x) is divided by (x + 1), the remainder is p(-1).
Thus, we have:
p(1) = (1⁴ - 2 × 1³ × + 3 × 1² - a × 1 + b)
= (1 - 2 + 3 - a + b)
= 2 - a + b
And,
p(-1) = [(-1)⁴ - 2 × (-1)³ + 3 × (-1)² - a × (-1) + b]
= (1 + 2 + 3 + a + b)
= 6 + a + b
Now,
2 - a + b = 5 ...(1)
6 + a + b = 19 ...(2)
Adding (1) and (2), we get:
8 + 2b = 24
⇒ 2b = 18
⇒ b = 8
By putting the value of b, we get the value of a, i.e., 5.
$\therefore$ a = 5 and b = 8
Now,
f(x) = x⁴ - 2x³ + 3x² - 5x + 8
Also,
When p(x) is divided by (x - 2), the remainder is p(2).
thus, we have:
p(2) = (2⁴ - 2 × 2³ + 3 × 2² - 5 × 2 + 8) [a = 5 and b = 8]
= (16 - 16 + 12 - 10 + 8)
= 10

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Question 115 Marks

If (x³ + ax² + bx + 6) has (x - 2) as a factor and leaves a remainder 3 when divided by (x - 3), find the values of a and b.

Answer

Let f(x) = (x³ + ax² + bx + 6)
Now, by remainder theorem, f(x) when divided by (x - 3) will leave a remainder as f(3).
⇒ So, f(3) = 3³ + a 3² + b 3 + 6 = 3
⇒ 27 + 9a + 3b + 6 = 3
⇒ 9 a + 3b + 33 = 3
⇒ 9a + 3b = 3 - 33
⇒ 9a + 3b = -30
⇒ 3a + b = -10 ...(i)
Given that (x - 2) is a factor of f(x).
By the Factor Theorem, (x - a) will be a factor of f(x) if f(a) = 0 and therefore f(2) = 0.
⇒ f(2) = 2³ + a 2² + b 2 + 6 = 0
⇒ 8 + 4a+ 2b + 6 = 0
⇒ 4a + 2b = -14
⇒ 2a + b = -7 ...(ii)
Subtracting (ii) from (i), we get,
⇒ a = -3
Substituting the value of a = -3 in (i), we get,
3(-3) + b = -10
⇒ -9 + b = -10
⇒ b = -10 + 9
⇒ b = -1
$\therefore$ a = -3 and b = -1.

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Question 125 Marks

What must be subtracted from (x⁴ + 2x³ - 2x² + 4x + 6) so that the result is exactly divisible by (x² + 2x - 3)?

Answer

Let p(x) = x⁴ + 2x³ - 2x² + 4x + 6 and q(x) = x² + 2x - 3.
When p(x) is divided by q(x), the remainder is a linear expression in x.
So, let r(x) = ax + b be subtracted from p(x) so that p(x) - r(x) is divided by q(x).
Let f(x) = p(x) - r(x) = p(x) - (ax + b)
= (x⁴ + 2x³ - 2x² + 4x + 6) - (ax + b)
= x⁴ + 2x³ - 2x² + (4 - a)x + 6 - b
We have,
q(x) = x² + 2x - 3
= x² + 3x - x - 3
= x(x + 3) - 1(x + 3)
= (x + 3)(x - 1)
Clearly, (x + 3) and (x - 1) are factors of q(x).
Therefore, f(x) will be divisible by q(x) if (x + 3) and (x - 1) are factors of f(x).
i.e., f(-3) = 0 and f(1) = 0
Consider, f(-3) = 0
⇒ (-3)⁴ + 2(-3)³ - 2(-3)² + (4 - a)(-3) + 6 - b = 0
⇒ 81 - 54 - 18 - 12 + 3a + 6 - b = 0
⇒ 3 + 3a - b = 0
⇒ 3a - b = -3 ...(i)
And, f(1) = 0
⇒ (1)⁴ + 2(1)³ - 2(1)² + (4 - a)(1) + 6 - b = 0
⇒ 1 + 2 - 2 + 4 - a + 6 - b = 0
⇒ 11 - a - b = 0
⇒ -a - b = -11 ...(ii)
Subtracting (ii) from (i), we get
4a = 8
⇒ a = 2
Substituting a = 2 in (i), we get
3(2) - b = -3
⇒ 6 - b = -3
⇒ b = 9
Putting the values of a and b in r(x) = ax + b, we get
r(x) = 2x + 9
Hence, p(x) is divisible by q(x), if r(x) = 2x + 9 is subtracted from it.

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