4 Marks Questions - Maths STD 9 Questions - Vidyadip

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Question 14 Marks

x⁴ - 2x³ - 7x² + 8x + 12.

Answer

Let f(x) = x⁴ - 2x³ - 7x² + 8x + 12
The factors of constant term in f(x) are $\pm1,\pm2,\pm3,\pm4,\pm6$ and $\pm12.$
We have,
f(1) = 1 - 2 - 7 + 8 + 12 = 12
⇒ (x - 1) is not a factor of f(x)
f(-1) = 1 + 2 - 7 - 8 + 12 = 0
⇒ (x + 1) is a factor of f(x)
f(2) = 16 - 16 - 28 - 16 + 12 = 0
⇒ (x - 2) is a factor of f(x)
f(-2) = 16 + 16 - 28 - 16 + 12 = 0
⇒ (x + 2) is a factor of f(x)
f(3) = 81 - 54 - 63 + 24 + 12 = 0
⇒ (x - 3) is a factor of f(x)
Since f(x) is a polynomial of degree 4. So, it cannot have more than 4 linear factors.
Thus, factors of f(x) are (x +1), (x - 2), (x + 2) and (x - 3).
Therefore,
f(x) = k(x + 1)(x + 2)(x - 2)(x - 3)
x⁴ - 2x³ - 7x² + 8x + 12 = k(x + 1)(x + 2)(x - 2)(x - 3) ...(1)
Putting x = 0 on both sides, we get,
12 = k(1)(2)(-2)(-3)
12 = 12k
k = 1
Substituting k = 1 in (1), we get,
x⁴ - 2x³ - 7x² + 8x + 12 = k(x + 1)(x + 2)(x - 2)(x - 3)

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Question 24 Marks

x⁴ + 10x³ + 35x² + 50x + 24.

Answer

Let f(x) = x⁴ + 10x³ + 35x² + 50x + 24
Now, putting x = -1, we get
f(-1) = (-1)⁴ + 10(-1)³ + 35(-1)² + 50(-1) + 24
= 1 - 10 + 35 - 50 + 24 = 60 - 60
= 0
Therefore, (x + 1) is a factor of polynomial f(x).
Now,
f(x) = x³(x + 1) + 9x²(x + 1) + 26(x + 1) + 24(x + 1)
= (x + 1)(x³ + 9x² + 26x + 24)
= (x + 1)g(x) ...(1)
Where g(x) = x³ + 9x² + 26x + 24
Putting x = -2, we get:
g(-2) = (-2)³ + 9(-2)² + 26x(-2) + 24
= -8 + 36 - 52 + 24 = 60 - 60
= 0
Therefore, (x + 2) is the factor of g(x).
Now,
g(x) = x²(x + 2) + 7x(x + 2) + 12(x + 2)
= (x + 2)(x² + 7x + 12)
= (x + 2)(x² + 4x + 3x + 12)
= (x + 2)(x + 3)(x + 4) ...(2)
From equation (1) and (2), we get:
f(x) = (x + 1)(x + 2)(x + 3)(x + 4)
Hence,
(x + 1), (x + 2), (x + 3) and (x + 4) are the factors of polynomial f(x).

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Question 34 Marks

What must be added to x³ - 3x² - 12x + 19 so that the result is exactly divisibly by x² + x - 6?

Answer

Let p(x) = x³ - 3x² - 12x + 19 and q(x) = x² + x - 6
When p(x) is divided q(x) then the remainder is a linear equation, r(x).
Let r(x) = ax + b when added to p(x) we get an expression which is divisible by q(x).
f(x) = p(x) + r(x)
f(x) = x³ - 3x² - 12x + 19 + ax + b
= x³ - 3x² - 12x + ax + 19 + b
= x³ - 3x² - x(12 - a) + 19 + b
Now,
q(x) = x² + x - 6 = x² + 3x - 2x - 6 = (x + 3)(x - 2)
Also,
f(x) is divisible q(x)
$\therefore$ f(-3) = 0, f(2) = 0
f(-3) = (-3)³ - 3(-3)² - 12(-3) + 19 + a(-3) + b = 0
⇒ -27 + 27 + 36 + 19 - 3a + b = 0
⇒ b = 3a + 27 + 27 - 36 - 19
⇒ b = 3a - 1 ...(1)
f(2) = 2³ - 3(2)² - 12(2) + 19 + 2a + b = 0
⇒ 8 - 12 - 24 + 19 + 2a + b = 0
⇒ b = 9 - 2a ...(2)
Equating (1) and (2)
3a - 1 = 9 - 2a
⇒ 5a = 10
⇒ a = 2
$\therefore$ b = 3a - 1 3(2) - 1 = 5 [Substituting b = 5, in equation 1]
Hence,
ax + b = 2x + 5

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Question 44 Marks

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
p(x) = x³ - 6x² + 11x - 6, x = 1, 2, 3

Answer

p(x) = x³ - 6x² + 11x - 6, x = 1, 2, 3
We know that,
p(x) = x³ - 6x² + 11x - 6
given that the values of x are 1, 2 , 3
substitute x = 1 in p(x)
p(1) = 1³- 6(1)² + 11(1) - 6
= 1 - (6 × 1) + 11 - 6
= 1 - 6 + 11 - 6
= 0
Now, substitute x = 2 in p(x)
P(2) = 2³ - 6(2)² + 11(2) - 6
= (2 × 3) - (6 × 4) + (11 × 2) - 6
= 8 - 24 - 22 - 6
= 0
Now, substitute x = 3 in p(x)
P(3) = 3³- 6(3)² + 11(3) - 6
= (3 × 3) - (6 × 9) + (11 × 3) - 6
= 27 - 54 + 33 - 6
= 0
Since, the result is 0 for x = 1, 2, 3 these are the roots of x³ - 6x² + 11x - 6

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Question 54 Marks

Using factor theorem, factorize the following polynomials:
y³ - 7y + 6

Answer

Let f(y) = y³ - 7y + 6
The factors of constant term in f(y) are $\pm1,\pm,2,\pm3$ and $\pm6.$
We have,
f(1) = 1 - 7 + 6 = 0
⇒ (y - 1) is a factor of f(y)
f(-1) = -1 + 7 + 6 = 12
⇒ (y + 1) is a factor of f(y)
f(2) = 8 - 14 + 6 = 0
⇒ (y - 2) is a factor of f(y)
f(-2) = -8 + 14 + 6 = 12
⇒ (y + 2) is not a factor of f(y)
f(3) = 27 - 21 + 6 = 12
⇒ (y - 3) is not a factor of f(y)
f(-3) = -27 + 21 + 6 = 0
⇒ (y + 3) is a factor of f(y)
Since f(y) is a polynomial of degree 3. So, it cannot have more than 3 linear factors.
Thus, factors of f(y) are (y - 1)(y - 2) and (y + 3).
Therefore,
f(y) = k(y - 1)(y - 2)(y + 3)
y³ - 7y + 6 = k(y - 1)(y - 2)(y + 3) ...(1)
Putting y = 0 on both sides, we get,
6 = k(-1)(-2)(3)
6 = 6k
k = 1
Substituting k = 1 in (1), we get,
y³ - 7y + 6 = (y - 1)(y - 2)(y + 3)

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Question 64 Marks

Using factor theorem, factorize the following polynomials:
y³ - 2y² - 29y - 42

Answer

Let f(y) = y³ - 2y² - 29y - 42 be the given polynomial.
Now, putting y = -2, we get
f(-2) = (-2)³ - 2(-2)² - 29(-2) - 42
= -8 - 8 + 58 - 42
= -58 + 58 = 0
Therefore, (y + 2) is a factor of polynomial f(y).
Now,
f(y) = y²(y + 2) + 4y(y + 2) - 2(y + 2)
= (y + 2)(y² - 4y - 21)
= (y + 2)(y² - 7y + 3y - 21)
= (y + 2)(y + 3)(y - 7)
Hence (y + 2), (y + 3) and (y - 7) are the factors of polynomial f(y).

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Question 74 Marks

Using factor theorem, factorize the following polynomials:
x⁴ - 7x³+ 9x² + 7x - 10

Answer

Let f(x) = x⁴ - 7x³+ 9x² + 7x - 10
The factors of constant term in f(x) are $\pm1,\pm2,\pm5$ and $\pm10.$
We have,
f(1) = 1 - 7 + 9 + 7 - 10 = 0
⇒ (x - 1) is a factor of f(x)
f(-1) = 1 + 7 + 9 - 7 - 10 = 0
⇒ (x + 1) is a factor of f(x)
f(2) = 16 - 56 + 36 + 14 - 10 = 0
⇒ (x - 2) is a factor of f(x)
f(-2) = 16 + 56 - 36 - 14 - 10 = 10
⇒ (x + 2) is not a factor of f(x)
f(5) = 625 - 875 + 225 + 35 - 10 = 0
⇒ (x - 5) is a factor of f(x)
Since f(x) is a polynomial of degree 4. So, it cannot have more than 4 linear factors.
Thus, factors of f(x) are (x - 1), (x + 1), (x - 2) and (x - 5).
Therefore,
f(x) = k(x - 1)(x + 1)(x - 2)(x - 5)
x⁴ - 7x³+ 9x² + 7x - 10 = k(x - 1)(x + 1)(x - 2)(x - 5) ...(1)
Putting x = 0 on both sides, we get,
-10 = k(-1)(1)(-2)(-5)
-10 = -10k
k = 1
Substituting k = 1 in (1), we get,
x⁴ - 7x³+ 9x² + 7x - 10 = (x - 1)(x + 1)(x - 2)(x - 5)

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Question 84 Marks

Using factor theorem, factorize the following polynomials:
x³ - 6x² + 3x + 10

Answer

Let x = 2
f(2) = 2³ + 6(2)² + 3(2) + 10 = 8 - 24 + 6 + 10 = 0
$\therefore$ x = 2 is a solution f(x)
i. e (x - 2) is a factor of f(x)

By division algorithm
x³ - 6x² + 3x + 10 = (x - 2)(x² - 4x - 5)
= (x - 2)(x² - 5x + x - 5)
= (x - 2)(x(x - 5) + 1(x - 5))
= (x - 2)(x - 5)(x + 1)
$\therefore$ x³ - 6x² + 3x + 10 = (x - 2)(x - 5)(x + 1)

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Question 94 Marks

Using factor theorem, factorize the following polynomials:
x³ + 6x² + 11x + 6

Answer

Let x = 1
$\text{f(1)}=1^3+6(2)^2+11(1)+6\neq0$
Let x = -1
f(-1) = (-1)³ + 6(-1)² + 11(-1) + 6 = 12 - 12 = 0
$\therefore$ x = -1 is a solution
⇒ x + 1 = 0
i. e (x + 1) is a factor of f(x)

By division algorithm
x³ + 6x² + 11x + 6 = (x + 1)(x² + 5x + 6)
= (x + 1)(x² + 2x + 3x + 6)
= (x + 1)(x(x + 2) + 3(x + 2))
= (x + 1)(x + 2)(x + 3)
$\therefore$ x³ + 6x² + 11x + 6 = (x + 1)(x + 2)(x + 3)

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Question 104 Marks

Using factor theorem, factorize the following polynomials:
x³ - 3x² - 9x - 5

Answer

Let p(x) = x³ - 3x² - 9x - 5
The factors of 5 are $\pm1,\pm5.$
By hit and trial method
p(-1) = (-1)³ - 3(-1)² - 9(-1) - 5
= -1 - 3 + 9 - 5 = 0
So x + 1 is a factor of this polynomial p(x).
Let us find the quotient while dividing x³ - 3x² - 9x - 5 by x + 1
By long division

Now,
Dividend = Divisor × Quotient + Remainder
$\therefore$ x³ - 3x² - 9x - 5 = (x + 1)(x² - 4x - 5) + 0
= (x + 1)(x² - 5x + x - 5)
= (x + 1)[x(x - 5) + 1(x - 5)]
= (x + 1)(x - 5)(x + 1)
= (x - 5)(x + 1)(x + 1)

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Question 114 Marks

Using factor theorem, factorize the following polynomials:
x³ - 2x² - x + 2

Answer

Let f(x) = x³ - 2x² - x + 2
The factors of constant term in f(x) are $\pm1,\pm2.$
We have
f(1) = 1 - 2 - 1 + 2 = 0
⇒ (x - 1) is a factor of f(x)
f(-1) = -1 - 2 + 1 + 2 = 0
⇒ (x + 1) is a factor of f(x)
f(2) = 8 - 8 - 2 + 2 = 0
⇒ (x - 2) is a factor of f(x)
Since f(x) is a polynomial of degree 3. So, it cannot have more than 3 linear factors.
Thus, factors of f(x) are (x - 1), (x + 1) and (x - 2).
Therefore,
f(x) = k(x - 1)(x + 1)(x - 2)
x³ - 2x² - x + 2 = k(x - 1)(x + 1)(x - 2) ...(1)
Putting x = 0 on both sides, we get,
2 = k(-1)(1)(-2)
2 = 2k
k = 1
Substituting k = 1 in (1), we get,
x³ - 2x² - x + 2 = (x - 1)(x + 1)(x - 2)

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Question 124 Marks

Using factor theorem, factorize the following polynomials:
x³ + 2x² - x - 2

Answer

Let x = 1
f(1) = 1³ + 2(1)² - 1 - 2 = 0
$\therefore$ x = 1 is a solution
⇒ x - 1 = 0
i. e (x - 1) is a factor of f(x)

By division algorithm
x³ + 2x² - x - 2 = (x - 1)(x² + 3x + 2)
= (x - 1)(x² + 2x + x + 2)
= (x - 1)(x(x + 2) + 1(x + 2))
= (x - 1)(x + 2)(x + 1)
$\therefore$ x³ + 2x² - x - 2 = (x - 1)(x + 2)(x + 1)

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Question 134 Marks

Using factor theorem, factorize the following polynomials:
x³ - 23x² + 142x - 120

Answer

Let p(x) = x³ - 23x² + 142x - 120
We shall now look for all the factors of -120. Some of these are $1,\pm2,\pm3,\pm,\pm4,\pm5,\pm6,\pm8\\\pm12,\pm15,\pm20,\pm24,\pm30,\pm,60$
By trial, we find that p(1) = 0.
So x - 1 is a factor of p(x).
Now, we see that x³ - 23x² + 142x - 120 = x³ - x²- 22x² + 22x +120x - 120
= x²(x - 1) - 22x(x - 1) + 120(x - 1)
= (x - 1)(x² - 22x + 120) [Taking (x - 1) common]
We could have also got this by dividing p(x) by x - 1.
Now x² - 22x + 120 can be factorised either by splitting the middle term or by using the factor theorem.
By splitting the middle term, we have:
x² - 22x + 120 = x² - 12x - 10x+ 120
= x(x - 12) - 10(x - 12)
= (x - 12)(x - 10)
So,
x³ - 23x² + 142x - 120 = (x - 1)(x - 10)(x - 12)

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Question 144 Marks

Using factor theorem, factorize the following polynomials:
x³ + 13x² + 32x + 20

Answer

Let p(x) = x³ + 13x²+ 32x + 20
The factors of 20 are $\pm1,\pm2,\pm4,\pm5\dots$
By hit and trial method
p(-1) = (-1)³ + 13(-1)² + 32(-1) + 20
= -1 + 13 - 32 + 20
= 33 - 33 = 0
As p(-1) is zero, so x + 1 is a factor of this polynomial p(x).
Let us find the quotient while dividing x³ + 13x² + 32x + 20 by (x + 1)
By long division

We know that
Dividend = Divisor × Quotient + Remainder
x³ + 13x² + 32x + 20 = (x + 1)(x² + 12x + 20) + 0
= (x + 1)(x² + 10x + 2x + 20)
= (x + 1)[x(x + 10) + 2(x + 10)]
= (x + 1)(x + 10)(x + 2)
= (x + 1)(x + 2)(x + 10)

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Question 154 Marks

Using factor theorem, factorize the following polynomials:
x³ - 10x² - 53x - 42

Answer

Let f(x) = x³ - 10x² - 53x - 42 be the given polynomial.
Now, putting x = -1, we get
f(-1) = (-1)³ - 10(-1)² - 53(-1) - 42
= -1 - 10 + 53 - 42
= -53 + 53 = 0
Therefore, (x + 1) is a factor of polynomial f(x).
Now,
f(x) = x²(x + 1) - 11x(x + 1) - 42(x + 1)
= (x + 1)(x² - 11x - 42)
= (x + 1)(x² - 14x + 3x - 42)
= (x + 1)(x + 3)(x - 14)
Hence (x + 1), (x + 3) and (x - 14) are the factors of polynomial f(x).

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Question 164 Marks

Using factor theorem, factorize the following polynomials:
3x³ - x² - 3x + 1

Answer

Let f(x) = 3x³ - x² - 3x + 1
The factor of the coefficient of x³ is 3. So, the possible rational roots of f(x) are $\pm1$ and $\pm\frac{1}{3}$
We have,
f(1) = 3 - 1 - 3 + 1 = 0
⇒ (x - 1) is a factor of f(x)
f(-1) = -3 - 1 + 3 + 1
⇒ (x + 1) is a factor of f(x)
So, (x - 1) and (x + 1) are factors of f(x)
⇒ (x - 1)(x + 1) is a also a factor of f(x)
⇒ x² - 1 is a factor of f(x).
Let us now divide f(x) = 3x³ - x² - 3x + 1 by x² - 1 to get the other factors of f(x).
By long division, we have:

Therefore,
3x³ - x² - 3x + 1 = (x² - 1)(3x - 1)
Now,
(x² - 1) = (x - 1)(x + 1)
Hence,
3x³ - x² - 3x + 1 = (x - 1)(x + 1)(3x - 1)

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Question 174 Marks

Using factor theorem, factorize the following polynomials:
2y³ + y² - 2y - 1

Answer

Let p(y) = 2y³ + y² - 2y - 1
By hit and trial method
p(1) = 2(1)³ + (1)² - 2(1) - 1
= 2 + 1 - 2 - 1 = 0
So, y -1 is a factor of this polynomial.

Now,
By long division method,
$\therefore$ 2y³ + y² - 2y - 1 = (y - 1)(2y² + 3y + 1)
= (y - 1)(2y² + 2y + y + 1)
= (y - 1)[2y(y + 1) + 1(y + 1)]
= (y - 1)(y + 1)(2y + 1)

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Question 184 Marks

Using factor theorem, factorize the following polynomials:
2y³ - 5y² - 19y + 42

Answer

Let f(y) = 2y³ - 5y² - 19y + 42 be the given polynomial.
Now, putting y = 2, we get
f(2) = 2(2)³ - 5(2)² - 19(2) + 42
= 16 - 20 - 38 + 42
= -58 + 58 = 0
Therefore, (y - 2) is a factor of polynomial f(y).
Now,
f(y) = 2y²(y - 2) - y(y - 2) - 21(y - 2)
= (y - 2)(2y² - y - 21)
= (y - 2)(2y² - 7y + 6y - 21)
= (y - 2)(y + 3)(2y - 7)
Hence (y - 2), (y + 3) and (2y - 7) are the factors of polynomial f(y).

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Question 194 Marks

Show that (x - 2), (x + 3) and (x - 4) are factors of x³ - 3x² - 10x + 24.

Answer

F(x) = x³ - 3x² - 10x + 24
Let,
x - 2 = 0
⇒ x = 2
f(2) = 2³ - 3(2)² - 10(2) + 24
= 8 - 12 - 20 + 24
= 32 - 32 = 0
Let,
x + 3 = 0
⇒ x = -3
f(-3) = (-3)³ - 3(-3)² - 10(-3) + 24
= -27 - 3(9) + 30 + 24
= -27 - 27 + 30 + 24
= -54 + 54 = 0
Let,
x - 4 = 0
⇒ x = 4
f(4) = 4³ - 3(4)² - 10(4) + 24
= 64 - 3(16) - 40 + 24
= 64 - 48 - 40 + 24
= 88 - 88 = 0
$\because$ f(2) = 0, f(-3) = 0, f(4) = 0
$\therefore$ By factore theoram (x - 2), (x + 3) and (x - 4) are factors of x³ - 3x² - 10x + 24.

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Question 204 Marks

In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
$\text{f(x)}=3\text{x}^4+2\text{x}^3-\frac{\text{x}^3}{3}-\frac{\text{x}}{9}+\frac{2}{27},$ $\text{g(x)}=\text{x}+\frac{2}{3}$

Answer

Here,
$\text{f(x)}=3\text{x}^4+2\text{x}^3-\frac{\text{x}^3}{3}-\frac{\text{x}}{9}+\frac{2}{27},$
$\text{g(x)}=\text{x}+\frac{2}{3}$
From, the remainder theorem when f(x) is divided by $\text{g(x)}=\text{x}-\Big(-\frac{2}{3}\Big),$ the remainder will be equal to $\text{f}\Big(-\frac{2}{3}\Big)$
Substitute the value of x in f(x)
$\text{f}\Big(-\frac{2}{3}\Big)=3\Big(-\frac{2}{3}\Big)^4+2\Big(-\frac{2}{3}\Big)^3\\-\frac{\Big(-\frac{2}{3}\Big)^3}{3}-\Bigg[\frac{\big(-\frac{2}{3}\big)}{9}+\frac{22}{7}\Big(\frac{2}{27}\Big)\Bigg]$
$=3\Big(\frac{16}{81}\Big)+2\Big(\frac{-8}{27}\Big)-\frac{4}{(9\times3)}-\Big(\frac{-2}{(9\times3)}\Big)+\frac{2}{27}$
$=\Big(\frac{16}{27}\Big)-\Big(\frac{16}{27}\Big)-\frac{4}{27}+\Big(\frac{2}{27}\Big)+\frac{2}{27}$
$=\Big(\frac{4}{27}\Big)-\Big(\frac{4}{27}\Big)$
$=0$
Therefore, the remainder is 0.

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Question 214 Marks

In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
f(x) = x⁴ - 3x² + 4, g(x) = x - 2

Answer

Here,
f(x) = x⁴ - 3x² + 4
g(x) = x - 2
From, the remainder theorem when f(x) is divided by g(x) = x - 2 the remainder will be equal to f(2)
Let, g(x) = 0
⇒ x - 2 = 0
⇒ x = 2
Substitute the value of x in f(x)
f(2) = 2⁴ - 3(2)² + 4
= 16 - (3 × 4) + 4
= 16 - 12 + 4
= 20 - 12
= 8
Therefore, the remainder is 8.

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Question 224 Marks

In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
f(x) = x³ + 4x² - 3x + 10, g(x) = x + 4

Answer

Here,
f(x) = x³ + 4x² - 3x + 10
g(x) = x + 4
From, the remainder theorem when f(x) is divided by g(x) = x - (-4) the remainder will be equal to f(-4)
Let, g(x) = 0
⇒ x + 4 = 0
⇒ x = -4
Substitute the value of x in f(x)
f(-4) = (-4)³ + 4(-4)²- 3(-4) + 10
= - 64 + (4 × 16) + 12 + 10
= - 64 + 64 + 12 + 10
= 12 + 10
= 22
Therefore, the remainder is 22.

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Question 234 Marks

In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
f(x) = 9x³ - 3x² + x - 5, $\text{g(x)}=\text{x}-\frac{2}{3}$

Answer

Here,

f(x) = 9x³ - 3x² + x - 5,

$\text{g(x)}=\text{x}-\frac{2}{3}$

From, the remainder theorem when f(x) is divided by $\text{g(x)}=\text{x}-\frac{2}{3}$ the remainder will be equal to $\text{f}\Big(\frac{2}{3}\Big)$

Let, g(x) = 0

$\Rightarrow\ \text{x}-\frac{2}{3}=0$

$\Rightarrow\ \text{x}=\frac{2}{3}$

Substitute the value of x in f(x)

$\text{f}\Big(\frac{2}{3}\Big)=9\Big(\frac{2}{3}\Big)-3\Big(\frac{2}{3}\Big)^2+\Big(\frac{2}{3}\Big)-5$

$=9\Big(\frac{8}{27}\Big)-3\Big(\frac{4}{9}\Big)+\frac{2}{3}-5$

$=\Big(\frac{8}{3}\Big)-\Big(\frac{4}{3}\Big)+\frac{2}{3}-5$

$=\frac{8-4+2-15}{3}$

$=\frac{10-19}{3}$

$=\frac{-9}{3}$

$=-3$

Therefore, the remainder is -3.

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Question 244 Marks

In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
f(x) = 4x⁴ - 3x³ - 2x² + x - 7, g(x) = x - 1

Answer

Here,
f(x) = 4x⁴ - 3x³ - 2x² + x -7
g(x) = x - 1
From, the remainder theorem when f(x) is divided by g(x) = x - (-1) the remainder will be equal to f(1)
Let, g(x) = 0
⇒ x - 1 = 0
⇒ x = 1
Substitute the value of x in f(x)
f(1) = 4(1)⁴- 3(1)³- 2(1)² + 1 - 7
= 4 - 3 - 2 + 1 - 7
= 5 - 12
= -7
Therefore, the remainder is 7.

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Question 254 Marks

In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
f(x) = 2x⁴ - 6x³ + 2x² - x + 2, g(x) = x + 2

Answer

Here,
f(x) = 2x⁴ - 6x³ + 2x² - x + 2
g(x) = x + 2
From, the remainder theorem when f(x) is divided by g(x) = x - (-2) the remainder will be equal to f(-2)
Let, g(x) = 0
⇒ x + 2 = 0
⇒ x = -2
Substitute the value of x in f(x)
f(-2) = 2(-2)⁴ - 6(-2)³ + 2(-2)² - (-2) + 2
= (2 × 16) - (6 × (-8)) + (2 × 4) + 2 + 2
= 32 + 48 + 8 + 2 + 2
= 92
Therefore, the remainder is 92.

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Question 264 Marks

In the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not:
f(x) = x³ - 6x² + 11x - 6, g(x) = x² - 3x + 2

Answer

g(x) = x² - 3x + 2
= x² - 2x - x + 2
= x(x - 2) - 1(x - 2)
= (x - 2)(x - 1)
Let g(x) = 0
⇒ (x - 2)(x - 1) = 0
If x - 2 = 0 ⇒ x = 2
If x - 1 = 0 ⇒ x = 1
f(2) = 2³ - 6(2)² + 11(2) - 6
= 8 - 24 + 22 - 6
= 30 - 30
= 0
f(1) = 1³ - 6(1)² + 11(1) - 6
= 1 - 6 + 11 - 6
= 0
$\because$ f(2) = 0 and f(1) = 0, by factor theorem, (x - 2) and (x - 1) both are factors of f(x).
Hence, (x - 2)(x - 1) is a factor of f(x).

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Question 274 Marks

If x = 0 and x = -1 are the roots of the polynomial f(x) = 2x³- 3x² + ax + b, Find the of a and b.

Answer

We know that, f(x) = 2x³- 3x² + ax + b
Given, the values of x are 0 and -1
Substitute x = 0 in f(x)
f(0) = 2(0)³- 3(0)² + a(0) + b
= 0 - 0 + 0 + b
= b ...(1)
Substitute x = (-1) in f(x)
f(-1) = 2(-1)³- 3(-1)² + a(-1) + b
= -2 - 3 - a + b
= -5 - a + b ...(2)
We need to equate equations 1 and 2 to zero
b = 0 and -5 - a + b = 0
since, the value of b is zero
substitute b = 0 in equation 2
⇒ -5 - a = -b
⇒ -5 - a = 0
a = -5
the values of a and b are -5 and 0 respectively.

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Question 284 Marks

If the polynomials ax³ + 3x² − 13 and 2x³ − 5x + a, when divided by (x - 2) leave the same remainder, Find the value of a.

Answer

Here,
The polynomials are:
f(x) = ax³ + 3x² − 13
p(x) = 2x³ − 5x + a
equate, x - 2 = 0
x = 2
substitute the value of x in f(x) and p(x)
f(2) = (2)³ + 3(2)² − 13
= 8a + 12 - 13
= 8a - 1 ...(1)
p(2) = 2(2)³ - 5(2) + a
= 16 - 10 + a
= 6 + a ...(2)
f(2) = p(2)
⇒ 8a - 1 = 6 + a
⇒ 8a - a = 6 + 1
⇒ 7a = 7
⇒ a = 1
The value of a = 1.

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Question 294 Marks

If the polynomial 2x³ + ax² + 3x - 5 and x³ + x² - 4x + a leave the same remainder when divided by x - 2, Find the value of a.

Answer

Given, the polymials are:
f(x) = 2x³ + ax² + 3x - 5
p(x) = x³ + x² - 4x + a
The remainders are f(2) and p(2) when f(x) and p(x) are divided by x - 2
We know that,
f(2) = p(2) (given in problem)
we need to calculate f(2) and p(2)
for, f(2)
substitute (x = 2) in f(x)
f(2) = 2(2)³ + a(2)²+ 3(2) - 5
= (2 × 8) + 4a + 6 - 5
= 16 + 4a + 1
= 4a + 17 ...(1)
for, p(2)
substitute (x = 2) in p(x)
p(2) = 2³ + 2²- 4(2) + a
= 8 + 4 - 8 + a
= 4 + a ...(2)
Since, f(2) = p(2)
Equate equation 1 and 2
⇒ 4a + 17 = 4 + a
⇒ 4a - a = 4 - 17
⇒ 3a = -13
$\Rightarrow\ \text{a}=\frac{-13}{3}$
The value of $\text{a}=\frac{-13}{3}.$

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Question 304 Marks

If x³ + ax² - bx + 10 is divisible by x² - 3x + 2, find the values of a and b.

Answer

Let,
p(x) = x³ + ax² - bx + 10
g(x) = x² - 3x + 2
Let g(x) = 0
⇒ x² - 3x + 2 = 0
⇒ x² - 2x - x + 2
⇒ x(x - 2) - 1(x - 2) =0
⇒ x - 2 = 0, x - 1 = 0
$\therefore$ x = 2, x = 1
Since x² - 3x + 2 is a factor of p(x)
$\therefore$ (x - 2)(x - 1) is a factor of p(x)
Hence,
p(2) = 0, p(1) = 0
p(2) = 2³ + a(2)² - b(2) + 10 = 0
⇒ 8 + 4a - 2b + 10 = 0
⇒ 4a - 2b + 18 = 0
⇒ 4a = -18 + 2b
$\Rightarrow\ \text{a}=\frac{-18+2\text{b}}{4}=\frac{2(-9+\text{b})}{4}=\frac{-9+\text{b}}{2}\dots(1)$
p(1) = 13 + a(1)2 - b(1) + 10 = 0
⇒ a - b + 11 = 0
⇒ a = b - 11 ...(2)
Equadting (1) and (2)
$\frac{-9+\text{b}}{2}=\text{b}-11$
⇒ -9 + b = 2b - 22
⇒ -9 + 22 = 2b - b
⇒ 13 = b
Subsitituting b = 13 in equation (2)
a = b - 11 = 13 - 11 = 2
$\therefore$ a = 2, b = 13

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Question 314 Marks

If $\text{x}=-\frac{1}{2}$ is zero of the polynomial p(x) = 8x³ - ax² - x + 2, Find the value of a.

Answer

We know that, p(x) = 8x³ - ax² - x + 2
Given that the value of $\text{x}=-\frac{1}{2}$
Substitute the value of x in f(x)
$\text{p}\Big(-\frac{1}{2}\Big)=8\Big(-\frac{1}{2}\Big)^3-\text{a}\Big(-\frac{1}{2}\Big)^2-\Big(-\frac{1}{2}\Big)+2$
$=-8\Big(\frac{1}{8}\Big)-\text{a}\Big(\frac{1}{4}\Big)+\Big(\frac{1}{2}\Big)+2$
$=-1-\Big(\frac{\text{a}}{4}+\frac{1}{2}+2\Big)$
$=1-\Big(\frac{\text{a}}{4}+\frac{1}{2}\Big)$
$=\frac{3}{2}-\frac{\text{a}}{4}$
To, find the value of a, equal $\text{p}\Big(-\frac{1}{2}\Big)$ to zero
$\text{p}\Big(-\frac{1}{2}\Big)=0$
$\frac{3}{2}-\frac{\text{a}}{4}=0$
On taking L.C.M
$\frac{6-\text{a}}{4}=0$
$\Rightarrow6-\text{a}=0$
$\Rightarrow\text{a}=6$

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Question 324 Marks

If both x + 1 and x - 1 are factors of ax³ + x² - 2x + b, find the values of a and b.

Answer

Let,
x + 1 = 0
x = -1
$\because$ (x + 1) is a factor of p(x) = ax³ + x² - 2x + b
$\therefore$ p(-1) = 0
p(-1) = a(-1)³ + (-1)² - 2(-1) + b = 0
⇒ -a + 1 + 2 + b = 0
⇒ -a + 3 + b = 0
⇒ a = 3 + b ...(1)
Let,
x - 1 = 0
x = 1
$\because$ (x - 1) is a factor of p(x)
$\therefore$ p(1) = 0
p(1) = a(1)³ + 1² -2(1) + b = 0
⇒ a + 1 - 2 + b = 0
⇒ a = -b + 1 ...(2)
Equating (1) and (2)
⇒ 3 + b = -b + 1
⇒ b + b = 1 - 3
⇒ 2b = -2
⇒ b = -1
Substituting b = -1 in equation (2)
a = -(-1) + 1 = 1 + 1 = 2
$\therefore$ a = 2, b = -1

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Question 334 Marks

Find the values of a and b, if x² - 4 is a factor of ax⁴ + 2x³ - 3x² + bx - 4.

Answer

Let g(x) = x² - 4 is, f(x) = ax⁴ + 2x³ - 3x² + bx - 4.
Let g(x) = 0
⇒ x² - 4 = 0
⇒ x² = 4,
$\Rightarrow \text{x}=\pm2$
Since (x² - 4) is a factor of f(x).
$\therefore$ f(2) = 0 and f(-2) = 0
f(2) = a(2)⁴ + 2(2)³ - 3(2)² + b(2) - 4 = 0
⇒ 16a + 16 - 12 + 2b - 4 = 0
⇒ 16a + 2b = 0
⇒ 16a = -2b
$\Rightarrow\ \text{a}=\frac{-2\text{b}}{16}=\frac{-\text{b}}{8}\dots(1)$
Also, f(-2) = 0
f(-2) = a(-2)⁴ + 2(-2)³ - 3(-2)² + b(-2) - 4 = 0
⇒ 16a - 16 - 12 - 2b - 4 = 0
⇒ 16a - 2b - 32 = 0
⇒ 16a = 2b + 32
$\Rightarrow\ \frac{2\text{b}+32}{16}\dots(2)$
Equating equations (1) and (2)
$\Rightarrow\ \frac{-\text{b}}{8}=\frac{2\text{b}+32}{16}$
$\Rightarrow\ \frac{-16\text{b}}{8}=2\text{b}+32$
⇒ -2b = 2b + 32
⇒ -4b = 32
⇒ b = -8
Substituting b = -8 in equation (1)
$\text{a}=\frac{-\text{b}}{8}=\frac{-(-8)}{8}=\frac{8}{8}=1$
$\therefore\text{a}=1,\text{b}=-8$

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Question 344 Marks

Find the value of a, if x + 2 is a factor of 4x⁴ + 2x³ - 3x² + 8x + 5a.

Answer

Let g(x) = x + 2, f(x) = 4x⁴ + 2x³ - 3x² + 8x + 5a.
Let g(x) = 0
⇒ x + 2 = 0
⇒ x = -2,
$\because$ g(x) is a factor of f(x)
$\therefore$ f(-2) = 0
f(-2) = 4(-2)⁴ + 2(-2)³ - 3(-2)² + 8(-2) + 5a = 0
⇒ 4(16) + 2(-8) - 3(4) + 8(-2) + 5a = 0
⇒ 64 - 16 - 12 - 16 + 5a = 0
⇒ 20 + 5a = 0
⇒ 5a = -20
$\Rightarrow\ \text{a}=\frac{-20}{5}=-4$
$\therefore\ \text{a}=-4$

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Question 354 Marks

Find the rational roots of the polynomial f(x) = 2x³ + x² - 7x - 6

Answer

Given that f(x) = 2x³ + x² - 7x - 6
f(x) is a cubic polynomial with an integer coefficient. If the rational root in the form of $\frac{\text{p}}{\text{q}},$ the values of p are limited to factors of 6 which are ±1, ±2, ±3, ±6 and the values of q are limited to the highest degree coefficient i.e 2 which are ±1, ±2
here, the possible rational roots are $±1, ±2, ±3, ±6, ±\frac{1}{2}, ±\frac{3}{2}$
Let, x = -1
f(-1) = 2(-1)³ +(-1)²- 7(-1) - 6
= -2 + 1 + 7 - 6
= -8 + 8
= 0
Let, x = 2
f(-2) = 2(2)³ + (2)²- 7(2) - 6
= (2 × 8) + 4 - 14 - 6
= 16 + 4 -14 - 6
= 20 - 20
= 0
Let, $\text{x}=-\frac{3}{2}$
$\text{f}=\Big(-\frac{3}{2}\Big)=2\Big(-\frac{3}{2}\Big)^3+\Big(-\frac{3}{2}\Big)^2-7\Big(-\frac{3}{2}\Big)-6$
$=2\Big(-\frac{27}{4}\Big)+\frac{9}{4}-7\Big(-\frac{3}{2}\Big)-6$
$=2\Big(-\frac{27}{4}\Big)+\frac{9}{4}-\Big(-\frac{21}{2}\Big)-6$
$=-6.75+2.25+10.5-6$
$=12.75-12.75$
$=0$
But from all the factors only -1, 2 and $-\frac{3}{2}$ gives the result as zero
So, the rational roots of 2x³ + x² - 7x - 6 are -1, 2 and $-\frac{3}{2}$

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Question 364 Marks

Find the integral roots of the polynomial f(x) = x³+ 6x² + 11x + 6

Answer

Given, that f(x) = x³ + 6x² + 11x + 6
Clearly we can say that, the polynomial f(x) with an integer coefficient and the highest degree term coefficient which is known as leading factor is 1.
So, the roots of f(x) are limited to integer factor of 6, they are ±1, ±2, ±3, ±6
Let x = -1
f(-1) = (-1)³ + 6(-1)² + 11(-1) + 6
= -1 + 6 - 11 + 6
= 0
Let x = -2
f(-2) = (-2)³ + 6(-2)² + 11(-2) + 6
= -8 - (6 × 4) - 22 + 6
= -8 + 24 - 22 + 6
= 0
Let x = -3
f(-3) = (-3)³ + 6(-3)² + 11(-3) + 6
= -27 - (6 × 9) - 33 + 6
= –27 + 54 - 33 + 6
= 0
But from all the given factors only -1, -2, -3 gives the result as zero.
So, the integral multiples of x³ + 6x² + 11x + 6 are -1, -2, -3.

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Question 374 Marks

Factorize the following polynomials:
x³ + 13x² + 31x - 45 given that x + 9 is a factor.

Answer

Let f(x) = x³ + 13x² + 31x - 45 be the given polynomial.
Therefore, (x + 9) is a factor of the polynomial f(x).
Now,
f(x) = x²(x + 9) + 4x(x + 9) - 5(x + 9)
= (x + 9)(x² + 4x - 5)
= (x + 9)(x² +5x - x - 5)
= (x + 9)(x - 1)(x + 5)
Hence (x - 1), (x + 5) and (x + 9) are the factors of polynomial f(x).

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Question 384 Marks

2x⁴ - 7x³ - 13x² + 63x - 45

Answer

Let f(x) = 2x⁴ - 7x³ - 13x² + 63x - 45 be the given polynomial.
Now, putting x = 1, we get
f(1) = 2(1)⁴ - 7(1)³ - 13(1)² + 63(1) - 45
= 2 - 7 - 13 + 63 - 45 = 65 - 65
= 0
Therefore, (x - 1) is a factor of polynomial f(x).
Now,
f(x) = 2x³(x - 1) - 5x²(x - 1) - 18x(x - 1) + 45(x - 1)
= (x - 1)(2x³ - 5x² - 18x + 45)
= (x - 1)g(x) ...(1)
Where g(x) = 2x³ - 5x² - 18x + 45
Putting x = 3, we get:
g(3) = 2(3)³ - 5(3)² - 18(3) + 45
= 54 - 45 - 54 + 45
= 0
Therefore, (x - 3) is the factor of g(x).
Now,
g(x) = 2x²(x - 3) + x(x - 3) - 15(x - 3)
= (x - 3)(2x² + x - 15)
= (x - 3)(2x² + 6x - 5x - 15)
= (x - 3)(x + 3)(2x - 5) ...(2)
From equation (1) and (2), we get:
f(x) = (x - 1)(x - 3)(x + 3)(2x - 5)
Hence,
(x - 1), (x - 3), (x + 3) and (2x - 5) are the factors of polynomial f(x).

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