3 Marks Question

Question 13 Marks

Factorise:
3x³ - x² - 3x + 1

Answer

Let p(x) = 3x³ - x² - 3x + 1

Constant term of p(x) = -1

$\therefore$ Factors of 1 are ±1

By trial, we find that p(1) = 0, so (x - 1) is a factor of p(x).

Now, we see that 3x³ - x² - 3x - 1

= 3x² - 3x² + 2x² - 2x - x + 1

= 3x²(x - 1) + 2x(x - 1) -1(x - 1)

= (x - 1)(3x² + 2x - 1)

Now, (3x² + 2x - 1) = 3x² - x² - 3x - 1 [bi spliting middle term]

= 3x²(x + 1) -1(x + 1) = (x + 1)(3x - 1)

$\therefore$ 3x² - x² - 3x - 1 = (x - 1)(x + 1)(3x - 1)

View full question & answer→

Question 23 Marks

Simplify (2x - 5y)³ - (2x + 5y)³.

Answer

(2x - 5y)³ - (2x + 5y)³

={(2x - 5y) - (2x + 5y)}{(2x - 5y)² + (2x - 5y)(2x + 5y) + (2x + 5y)²}

$\Big[\therefore$ a³ - b³ = (a - b)(a² + ab + b²)$\Big]$

= (2x - 5y - 2x - 5y)(4x² + 25y² - 20xy + 4x² - 25y² + 4x² + 25y² + 20xy)

= (-10y)(2x² + 25y²)

= -120x²y - 250y³

View full question & answer→

Question 33 Marks

Factorise:
x³ - 6x² + 11x - 6

Answer

Let p(x) = x³ - 6x² + 11x - 6
Constant term of p(x) = -6
$\therefore$ Factors of -6 are ±1, ±2, ±3, ±5, ±6
By trial, we find that p(1) = 0, so (x - 1) is a factor of p(x).
$[\therefore$ (1)³ - 6(1)² + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$]$
Now, we see that x³ - 6x² + 11x - 6
= x³ - x² - 5x² + 5x + 6x - 6
= x² (x - 1) - 5x(x - 1) + (6x - 1)
= (x - 1)(x² - 5x + 6) [taking (x - 1) common factor]
Now, (x² - 5x + 6) = x² - 3x - 2x + 6 [by spliting middle term]
= x(x - 3) -2(x - 3)
= (x - 3)(x - 2)
$\therefore$ x³ - 6x² + 11x - 6 = (x - 1)(x - 2)(x - 3)

View full question & answer→

Question 43 Marks

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

Answer

Let f(x) = p(x)² + 5x + r
Since, x - 2 is a factor of f(x), then f(2) = 0
$\therefore$ p(2)² + 5(2) + r = 0
⇒ 4p + 10 + r = 0 ...(i)
Since, $\text{x}-\frac{1}{2}$ is a factor of f(x), then $\text{f}\Big(\frac{1}{2}\Big)=0$
$\therefore\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
$\Rightarrow\text{p}\times\frac{1}{4}+\frac{5}{2}+\text{r}=0$
$\Rightarrow\text{p}+10+4\text{r}=0\ ...\text{(i)}$
Since, x - 2 and $\text{x}-\frac{1}{2}$ is a factors of f(x) = px²+ 5x + r.
From Eqs. (i) and (ii), 4p + 10 + r = p + 10 + 4r
⇒ 3p = 3r
$\therefore$ p = r

View full question & answer→

Question 53 Marks

Find the value of the polynomial 3x³ - 4x² + 7x - 5, when x = 3 and also when x = -3.

Answer

Let p(x) = 3x² - 4x² + 7x - 5
$\therefore $ p(3) = 3(3)² - 4(3)² + 7(3) - 5
= 3(27) - 4(9) + 21 - 5
= 81 - 36 + 21 - 5
= 61
Now, p(-3) = 3(-3)² - 4(-3)² + 7(-3) - 5
= 3(-27) - 4(9) - 21 - 5
= -81 - 36 - 21 - 5
= -143

View full question & answer→

Question 63 Marks

Find the following product:
(2x - y + 3z) (4x² + y² + 9z² + 2xy + 3yz - 6xz)

Answer

(2x - y + 3z) (4x² + y² + 9z² + 2xy + 3yz - 6xz)

= {2x + (-y) + 3z}{(2x)² + (-y)² + (3z)² - (2x)(-y) - (-y)(3z) - (3z)(2x)}

= (2x)³ + (-y)³ + (3z)³ - 3(2x)(-y)(3z)

$\big[\therefore$ (a + b + c)(a² + b² + c² - ab - bc - ca) = a³ + b³ + c³ - 3abc $\big]$

= 8x³ - y³ + 27z² + 18xyz

View full question & answer→

Question 73 Marks

Determine which of the following polynomials has x - 2 a factor:

3x² + 6x - 24
4x²+ x – 2

Answer

Given

x - 2 = 0

⇒ x = 2

Now, we have two equations

3x² + 6x – 24

Put x = 2 in this equation

3(2)² + 6 × 2 - 24

3 × 4 + 12 - 24

12 + 12 - 24 = 0

X - 2 is a polynomial factor for this equation.

4x²+ x – 2

Put x = 2 in this equation

4(2)² + 2 - 2

4 × 4 + 2 - 2 = 16

X - 2 is a not a polynomial factor for this equation.

Hence, the equation 3x² + 6x – 24 is an equation which has a polynomial factor of x - 2.

View full question & answer→

Question 83 Marks

For the polynomial $\frac{\text{x}^3+2\text{x}+1}{5}-\frac{7}{2}\text{x}^2-\text{x}^6,$ write.

The degree of the polynomial.
The coefficient of x³.
The coefficient of x^6.
The constant term.

Answer

$\frac{\text{x}^3+2\text{x}+1}{5}-\frac{7}{2}\text{x}^2-\text{x}^6$
$\frac{\text{x}^3}{5}+\frac{2\text{x}}{5}+\frac{1}{5}-\frac{7}{2}\text{x}^2-\text{x}^6$

We know that highest power of variable in a polynomial is the degree of the polynomial. In the given polynomial, the term with highest of x is - x, and the exponent of x in this term in 6.
The coefficient of x³ is $\frac{1}{5}$
The coefficient of x⁶is -1.
The constant term is $\frac{1}{5}$

View full question & answer→

Question 93 Marks

Check whether p(x) is a multiple of g(x) or not:
p(x) = 2x³ - 11x² - 4x + 5, g(x) = 2x + 1

Answer

p(x) will be a multiple g(x) if g(x) divides p(x).
Now, g(x) = 2x + 1 give $\text{x}=-\frac{1}{2}$
Remainder $=\text{p}\Big(-\frac{1}{2}\Big) = 2\Big(\frac{-1}{2}\Big) ^3 - 11\Big(\frac{-1}{2}\Big) ^2 - 4\Big(\frac{-1}{2}\Big) +5$
$=2\Big(\frac{-1}{8}\Big)-11\Big(\frac{1}{4}\Big)+2+5=\frac{-1}{4}-\frac{11}{4}+7$
$=\frac{-1-11+28}{4}=\frac{16}{4}=4$
Since remainder ≠ 0, So p(x) is not a multiple of g(x).

View full question & answer→

Question 103 Marks

Show that:
2x - 3 is a factor of x + 2x³ - 9x² + 12.

Answer

Let p(x) = x + 2x³ - 9x² + 12, g(x) = 2x - 3
g(x) = 2x - 3 = 0 gives $\text{x}=\frac{3}{2}$
g(x) will be a factor of p(x) if $\text{p}\Big(\frac{3}{2}\Big)=0$ (Factor theorem)
Now, $\text{p}\Big(\frac{3}{2}\Big)=\frac{3}{2}+2\Big(\frac{3}{2}\Big)^3-9\Big(\frac{3}{2}\Big)^2+12$
$=\frac{3}{2}+2\Big(\frac{27}{8}\Big)-9\Big(\frac{9}{4}\Big)+12$
$=\frac{3}{2}+\frac{27}{4}-\frac{81}{4}+12$
$=\frac{6+27-81+48}{4}=\frac{0}{4}=0$
Since, $\text{p}\Big(\frac{3}{2}\Big)=0,$ So g(x) is a factor of p(x).

View full question & answer→

Question 113 Marks

If x + 1 is a factor of ax³ + x² - 2x + 4a - 9, find the value of a.

Answer

Let p(x) = ax³ + x² - 2x + 4a - 9
As (x + 1) is a factor of p(x)
$\therefore$ p(-1) = 0 [By factor theorem]
⇒ a(-1)³ + (-1)² - 2(-1) + 4a - 9 = 0
⇒ a(-1) + 1 + 2 + 4a - 9 = 0
⇒ -a + 4a - 6 = 0
⇒ 3a - 6 = 0 ⇒ 3a = 6 ⇒ a = 2

View full question & answer→

Question 123 Marks

Factorise:
2x³ - 3x² - 17x + 30

Answer

Let p(x) = 2x³ - 3x² - 17x + 30
Constant term of p(x) = 30
$\therefore$ Factors of 30 are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30
By trial, we find that p(2) = 0, so (x - 2) is a factor of p(x).
$[\therefore$ 2(2)³ - 3(2)² - 17(2) + 30 = 16 - 12 - 34 + 30 = 0$]$
Now, we see that 2x³ - 3x² - 17x + 30
= 2x³ - 4x² + x² - 2x - 15x + 30
= 2x²(x - 2) + x(x - 2) - 15(x - 2)
=(x - 2)(2x² + x - 15) [taking(x - 2) common factor]
Now, (2x² + x - 15) can be factorised either by spliting the middle term or by using the factor theorem.
Now, (2x² + x - 15) = 2x² + 6x - 5x - 15
= 2x(x + 3) -5(x + 3) [by spliting the middle term]
= (x + 3)(2x - 5)
$\therefore$ 2x³ - 3x² - 17x + 30 = (x - 2)(x + 3)(2x - 5)

View full question & answer→

Question 133 Marks

Find the value of m so that 2x - 1 be a factor of 8x⁴ + 4x³ - 16x²+ 10x + m.

Answer

Let p(x) = 8x⁴ + 4x³ - 16x²+ 10x + m
Since, 2x - 1 is a factor of p(x), then put $\text{p}\Big(\frac{1}{2}\Big)=0$
$\therefore8\Big(\frac{1}{2}\Big)^4+4\Big(\frac{1}{2}\Big)^3-16\Big(\frac{1}{2}\Big)^2+10\Big(\frac{1}{2}\Big)+\text{m}=0$
$\Rightarrow8\times\frac{1}{16}+4\times\frac{1}{8}-16\times\frac{1}{4}+10\Big(\frac{1}{2}\Big)+\text{m}=0$
$\Rightarrow\frac{1}{2}+\frac{1}{2}-4+5+\text{m}=0$
$\Rightarrow1+1+\text{m}=0$
$\therefore\text{m}=-2$
Hence, the value of m is -2.

View full question & answer→

Question 143 Marks

If the polynomials az³ + 4z² + 3^z - 4 and z³ - 4z + a leave the same remainder when divided by z - 3, find the value of a.

Answer

p(z) = az³ + 4z² + 3^z - 4
p(z) = z³ - 4z + a
As these two polynomials leave the same remainder, when divided by z - 3, then p(3) = q(3).
$\therefore$ p(3) = a(3)³ + 4(3)² + 3(3) - 4
= 27a + 36 + 9 - 4
or p(3) = 27a + 41
And q(3) = (3)³ - 4(3) + a
= 27 - 12 + a = 15 + 1
Now, p(3) = q(3)
⇒ 27a + 41 = 15 + a
⇒ 26a = -26a, a = -1
Hence, the required value of a = -1.

View full question & answer→

Question 153 Marks

Show that p - 1 is a factor of p¹⁰ - 1 and also of p¹¹ - 1.

Answer

If p - 1 is a factor of p¹⁰ - 1, then (1)¹⁰ - 1 should be equal to zero.
Now, (1)¹⁰ - 1 = 1 - 1 = 0
Therefore, p - 1 is a factor of p¹⁰ - 1.
Again, if p - 1 is a factor of p¹¹ - 1, then (1)¹¹ - 1 should be equal to zero.
Now, (1)¹¹ - 1 = 1 - 1 = 0
Therefore, p - 1 is a factor of p¹¹ - 1.
Hence, p - 1 is a factor of p¹⁰ - 1 and also of p¹¹ - 1.

View full question & answer→

Question 163 Marks

If x + 2a is a factor of x⁵ – 4a² x³ + 2x + 2a + 3, find a.

Answer

Let p(x) = x⁵ – 4a² x³ + 2x + 2a + 3
If x - (-2a) is a factor of p(x), then p(-2a) = 0
$\therefore$ p(-2a) = (-2a)⁵ - 4a²(-2a)³ + 2(-2a) + 2a + 3
= -32a⁵ + 32a⁵ - 4a + 2a + 3
= -2a + 3
Now, p(-2a) = 0
⇒ -2a + 3 = 0
$\Rightarrow\text{a}=\frac{3}{2}$

View full question & answer→

Question 173 Marks

If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ - 3abc = -25.

Answer

To prove, a³ + b³+ c³ - 3abc = -25
Given,
a + b + c = 5, ab + bc + ca = 10
$\therefore$ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
$\therefore$ (5)² = a² + b² + c² = 25(10)
⇒ 25 = a² + b² + c² = 20
⇒ a² + b² + c² = 25 - 20
⇒ a² + b² + c² = 5
LHS = a³ + b³+ c³ - 3abc
= (a + b + c)(a² + b² + c² - ab - bc - ca)
= (5)[5 - (ab + bc + ca)]
= 5(5 - 10) = 5(-5) = -25 = RHS
Hence proved.

View full question & answer→

Question 183 Marks

Factorise:
x³ + x² - 4x - 4

Answer

Let p(x) = x³ + x² - 4x - 4
Constant term of p(x) = -4
$\therefore$ Factors of -4 are ±1, ±2, ±4
By trial, we find that p(-1) = 0, so (x + 1) is a factor of p(x).
Now, we see that x³ + x² - 4x - 4
= x²(x + 1) - 4(x + 1)
= (x + 1)(x² - 4) [taking (x + 1) common factor]
Now, x² - 4 = x²- 2²
= (x + 2)(x - 2) [Using identity, a² - b² =(a - b)(a + b)]
$\therefore$ x³ + x² - 4x - 4 = (x + 1)(x - 2)(x + 2)

View full question & answer→

Question 193 Marks

If a + b + c = 9 and ab + bc + ca = 26, find a² + b² + c².

Answer

We have that (a + b + c)² = a² + b² + c² + 2ab + bc + 2ca
⇒ (a + b + c)² = (a² + b² + c²) + 2(ab + bc + ca)
⇒ 9² = (a² + b² + c²) + 2(26)
[Putting the value of a + b + c and ab + bc + ca]
⇒ 81 = (a² + b² + c²) + 52
⇒ (a² + b² + c²) = 81 - 52 = 29

View full question & answer→

Maths 3 Marks Question

Test yourself on this topic