Introduction to Linear Polynomials

2x² − 5x + 3

Highest power of the variable = 2.

y³ + 2y − 1

Highest power of the variable = 3.

−9 is a nonzero constant.

−9 = −9x⁰

4z − 3

Highest power of the variable = 1.

Degree 1: x + 1

Degree 2: x² + 1

Degree 3: x³ + 1

x⁴ − 3x³ + 6x² − 2x + 7

x² term = 6x²

x³ term = −3x³

4z³ + 5z² − 11

= 4z³ + 5z² + 0z − 11

The term without x is −10.

5x − 3

= 5(0) − 3

= 0 − 3

= −3

5x − 3

= 5(−1) − 3

= −5 − 3

= −8

5x − 3

= 5(2) − 3

= 10 − 3

= 7

7s² − 4s + 6

= 7(0)² − 4(0) + 6

= 0 − 0 + 6

= 6

7s² − 4s + 6

= 7(−3)² − 4(−3) + 6

= 7 × 9 + 12 + 6

= 63 + 12 + 6

= 81

7s² − 4s + 6

= 7(4)² − 4(4) + 6

= 112 − 16 + 6

= 102

Let the child’s present age be x years; the mother’s age is 3x.

(x + 5) + (3x + 5) = 70

4x + 10 = 70

4x = 60

x = 15

3x = 45

Let the integers be 2k and 5k.

5k − 2k = 63

3k = 63

k = 21

2k = 42; 5k = 105

Let x be the number of ₹5 coins. Then there are 3x coins of ₹2.

5x + 2(3x) = 88

11x = 88

x = 8

3x = 24

Let the shorter length be x ft; the longer is 4x ft.

x + 4x = 300

5x = 300

x = 60

4x = 240

Let width = w cm and length = 2w + 3 cm.

2[(2w + 3) + w] = 24

3w + 3 = 12

3w = 9

w = 3

Length = 2(3) + 3 = 9 cm

Let n be the number of completed months.

A(n) = 500 + 150n

A(2) = 500 + 300 = ₹800

A(3) = 500 + 450 = ₹950

A(4) = 500 + 600 = ₹1,100

Let n be the number of completed hours.

M(n) = 120 − 9n

M(1) = 111; M(2) = 102; M(3) = 93

Use this fixed-loss rule only while the count stays nonnegative.

A = length × breadth

= 13 × 12

= 156 cm²

A = length × breadth

= 13 × 10

= 130 cm²

A = length × breadth

= 13 × 8

= 104 cm²

A = 13 × b

= 13b

V = length × breadth × height

= 7 × 11 × 5

= 77 × 5

= 385 cm³

V = length × breadth × height

= 7 × 11 × 9

= 77 × 9

= 693 cm³

V = length × breadth × height

= 7 × 11 × 13

= 77 × 13

= 1,001 cm³

V = 7 × 11 × h

= 77h

Pages left after n days = 500 − 20n

After 15 days: 500 − 20(15)

= 500 − 300

= 200

h = 1.75 + 0.5(7)

= 1.75 + 3.5

= 5.25 ft

h(t) = 1.75 + 0.5t

Substitute t = 0, 1, …, 10 to complete the table.

h(t) = 1.75 + 0.5t

A constant increase of 0.5 ft per month gives linear growth.

v = 10,000 − 800(3)

= 10,000 − 2,400

= ₹7,600

v(t) = 10,000 − 800t

Substitute t = 0, 1, …, 8 to complete the table.

v(t) = 10,000 − 800t

A constant decrease of ₹800 per year gives linear decay.

P = 750 + 50(6)

= 750 + 300

= 1,050

P(t) = 750 + 50t

Substitute t = 0, 1, …, 10 to complete the table.

P(t) = 750 + 50t

A constant increase of 50 people per year gives linear growth.

b(x) = 600 − 15x

The balance decreases by ₹15 per day: linear decay.

600 − 15x = 0

15x = 600

x = 600 ÷ 15

= 40

b(x) = 600 − 15x

Substitute x = 1, 2, …, 10 to complete the table.

10a + b = 400

14a + b = 500

4a = 100 ⇒ a = 25

b = 400 − 10(25) = 150

y = 25x + 150

10a + b = 800

15a + b = 1,100

5a = 300 ⇒ a = 60

b = 800 − 10(60) = 200

y = 60x + 200

32a + b = 0

212a + b = 100

180a = 100 ⇒ a = 5/9

b = −32(5/9) = −160/9

C = (5/9)(F − 32)

Plot the points from the table and join each set with a straight line.

y = 4x

y = 2x

y = x

Plot the points from the table and join each set with a straight line.

y = −6x

y = −3x

y = −x

Plot the points from the table and join each set with a straight line.

y = 5x

y = −5x

Plot the points from the table and join each set with a straight line.

y = 3x − 1

y = 3x

y = 3x + 1

Plot the points from the table and join each set with a straight line.

y = −2x − 3

y = −2x

y = 2x + 3

x³ − 7x² + 1

Degree = 3; coefficient of x² = −7.

5x² − 3x + 7

= 5(1)² − 3(1) + 7

= 5 − 3 + 7

= 9

4t³ − t² + 6

= 4a³ − a² + 6

Let the number be x.

(5/2)x + 2/3 = −7/12

(5/2)x = −7/12 − 8/12

= −15/12 = −5/4

x = (−5/4) × (2/5)

= −1/2

Let the smaller number be x; the larger is 5x.

5x + 21 = 2(x + 21)

5x + 21 = 2x + 42

3x = 21

x = 7

5x = 35

A(n) = 800 + 250n

A(6) = 800 + 250(6)

= 800 + 1,500

= ₹2,300

2 years = 24 months

A(n) = 800 + 250n

A(24) = 800 + 250(24)

= 800 + 6,000

= ₹6,800

Let a be the larger digit and b the smaller digit.

(10a + b) + (10b + a) = 143

11(a + b) = 143 ⇒ a + b = 13

a − b = 3

2a = 16 ⇒ a = 8

b = 13 − 8 = 5

y = −3x + 4

Slope = −3; y-intercept = 4

At x = 0: y = 4

2y = 4x + 7

y = 2x + 7/2

Slope = 2; y-intercept = 7/2

5y = 6x − 10

y = (6/5)x − 2

Slope = 6/5; y-intercept = −2

3y = 6x − 11

y = 2x − 11/3

Slope = 2; y-intercept = −11/3

Slopes: −3, 2, 6/5, 2

Lines (ii) and (iv) have the same slope, 2, and different y-intercepts.

F = (9/5)(K − 273) + 32

= (9/5)(313 − 273) + 32

= (9/5)(40) + 32

= 72 + 32

= 104 °F

158 = (9/5)(K − 273) + 32

126 = (9/5)(K − 273)

K − 273 = 126 × (5/9) = 70

K = 70 + 273

= 343 K

w = force × distance

= 3d

At d = 2: w = 3(2) = 6

Plot the table points and join them to draw w = 3d.

p(x) = ax + b

a + b = 5

3a + b = 11

2a = 6 ⇒ a = 3

b = 5 − 3 = 2

At x = 0: y = 3(0) + 2 = 2

At y = 0: 3x + 2 = 0

3x = −2 ⇒ x = −2/3

y = 3x + 2

At x = 1: y = 5

At x = 3: y = 11

Plot the table points and join them with a straight line.

p(0) = 5 ⇒ b = 5

p(x) + q(x) = 6x + 4

a + c = 6; b + d = 4 ⇒ d = −1

(p − q)(3) = 0

3(a − c) + (5 − (−1)) = 0

a − c = −2

2a = 4 ⇒ a = 2; c = 4

Stage 4: 6 + 5(4 − 1) = 21

Stage 5: 6 + 5(5 − 1) = 26

T(n) = 6 + 5(n − 1)

= 5n + 1

Each new hexagon adds 5 sticks.

First hexagon: 6 sticks. Each of the remaining n − 1 adds 5.

T(n) = 6 + 5(n − 1)

= 6 + 5n − 5

= 5n + 1

T(15) = 5(15) + 1

= 75 + 1

= 76

5n + 1 = 200

5n = 199

n = 199/5 = 39.8

A stage number must be a positive integer.

2a + b = 3; 6a + b = 11

4a = 8 ⇒ a = 2

b = 3 − 2(2) = −1

p(x) = 2x − 1

Parallel lines have the same slope, so c = 2.

−1 = 2(4) + d ⇒ d = −9

q(x) = 2x − 9

p(x) = 0 ⇒ x = 1/2

q(x) = 0 ⇒ x = 9/2

f(x) = ax + a = a(x + 1)

At x = −1: f(−1) = a(−1 + 1) = 0

Slope = a > 0

200l + 160w + 50lw

Terms: 200l, 160w, 50lw.

Variables: l and w.

200l + 160w + 50lw: two variables, including the product lw.

4x + 5y + 3: two separate variable terms and the constant 3.

10x − x²

Terms: 10x and −x². Variable: x.

10x − x²: one variable; highest power 2.

4x + 5y + 3: two variables, each with power 1; constant 3.

P = 4s

For s = 1, 1.5, 2, 2.5, 3: P = 4, 6, 8, 10, 12

Increase in P = 4 × 0.5 = 2 cm

Let m be the number of matches.

200 + 50m = 750

50m = 550

m = 11

2(length + width) = 20

Width = 10 − x

A(x) = x(10 − x) = 10x − x²

A(6) = 10(6) − 6²

= 60 − 36

= 24 cm²

1, 3, 5, 7, …

Add 2 tiles at each stage.

1, 3, 5, 7, 9, 11, 13

T(15) = 2(15) − 1

= 30 − 1

= 29

T(26) = 2(26) − 1

= 52 − 1

= 51

2n − 1 = 21

2n = 22

n = 11

2n − 1 = 47

2n = 48

n = 24

Money left = 100 − 5(15)

= 100 − 75

= ₹25

100 − 5n = 0

5n = 100

n = 20

Let n be the distance in km (n ≥ 2).

25 + 15(n − 2) = 130

15(n − 2) = 105

n − 2 = 7

n = 9

C(15) = 100 + 60(15)

= 100 + 900

= ₹1,000

100 + 60d = 700

60d = 600

d = 10

h(5) = 3 − 0.5(5)

= 3 − 2.5

= 0.5 m

y = 20x + 150

20x: data charge at ₹20 per GB.

150: fixed monthly fee of ₹150.

y = 2x + 1

Substitute each x-value to fill the table.

Plot each (x, y) pair, then join the points with a straight line.

y = ax; a > 0

At x = 0, y = 0: every line passes through the origin.

a > 1: steeper than y = x. 0 < a < 1: less steep.

y = −ax; a > 0

At x = 0, y = 0: every line passes through the origin.

a > 1: steeper fall than y = −x. 0 < a < 1: gentler fall.

y = 3x + 1: slope = 3

y = −3x + 1: slope = −3

At x = 0: y = 1 for both

y = ax + b

Same a: same slope. Different b: different y-axis crossings.