Chapter 2: Introduction to Linear Polynomials (Linear Polynomials)

The degree is the highest power of the variable.

• (i) $2x^2-5x+3$: highest power of x is 2, so the degree is 2 (quadratic).
• (ii) $y^3+2y-1$: highest power of y is 3, so the degree is 3 (cubic).
• (iii) $-9=-9x^0$: a non-zero constant has degree 0.
• (iv) $4z-3$: highest power of z is 1, so the degree is 1 (linear).

(i) 2 (ii) 3 (iii) 0 (iv) 1

Any polynomial whose highest power is the required degree will do. For example:

• Degree 1: $3x+2$
• Degree 2: $x^2-4x+1$
• Degree 3: $2y^3+y^2-5$

For example: $3x+2$ (degree 1), $x^2-4x+1$ (degree 2), $2y^3+y^2-5$ (degree 3).

The coefficient is the number multiplying that power, taken with its sign. The term with $x^3$ is $-3x^3$ and the term with $x^2$ is $+6x^2$.

Coefficient of $x^2$ = 6; coefficient of $x^3$ = −3.

There is no term containing $z$ (to the power 1). We can write the polynomial as $4z^3+5z^2+0\cdot z-11$.

The coefficient of z is 0.

The constant term is the term without any variable, taken with its sign.

The constant term is −10.

$$\text{(i) }5(0)-3=-3\text{(ii) }5(-1)-3=-5-3=-8\text{(iii) }5(2)-3=10-3=7$$

(i) −3 (ii) −8 (iii) 7

$$\text{(i) }7(0{)}^2-4(0)+6=6$$

$$\text{(ii) }7(-3{)}^2-4(-3)+6=7\times 9+12+6=63+18=81$$

$$\text{(iii) }7(4{)}^2-4(4)+6=112-16+6=102$$

(i) 6 (ii) 81 (iii) 102

Let Salil's present age be $x$ years. Then his mother's present age is $3x$ years. After 5 years, their ages will be $x+5$ and $3x+5$.

$$(x+5)+(3x+5)=70\Rightarrow 4x+10=70\Rightarrow 4x=60\Rightarrow x=15$$

Mother's age $=3\times 15=45$ years. Check: after 5 years, $20+50=70$ ✓

Salil is 15 years old and his mother is 45 years old.

Since the ratio is 2 : 5, let the integers be $2k$ and $5k$ for some positive number $k$.

$$5k-2k=63\Rightarrow 3k=63\Rightarrow k=21$$

The integers are $2\times 21=42$ and $5\times 21=105$. Check: $105-42=63$ and $42:105=2:5$ ✓

The integers are 42 and 105.

Let the number of five-rupee coins be $x$. Then she has $3x$ two-rupee coins.

$$5\times x+2\times 3x=88\Rightarrow 5x+6x=88\Rightarrow 11x=88\Rightarrow x=8$$

Two-rupee coins $=3\times 8=24$. Check: $8\times 5+24\times 2=40+48=88$ ✓

8 five-rupee coins and 24 two-rupee coins.

Let the shorter piece be $x$ feet. The longer piece is $4x$ feet.

$$x+4x=300\Rightarrow 5x=300\Rightarrow x=60$$

Longer piece $=4\times 60=240$ feet. Check: $60+240=300$ ✓

The pieces are 60 feet and 240 feet long.

Let the width be $w$ cm. Then the length is $2w+3$ cm.

$$2[(2w+3)+w]=24\Rightarrow 3w+3=12\Rightarrow 3w=9\Rightarrow w=3$$

Length $=2\times 3+3=9$ cm. Check: $2(9+3)=24$ ✓

Length = 9 cm, width = 3 cm.

Each month the amount increases by the same ₹150, starting from ₹500.

From the second month onwards she has ₹800, ₹950, ₹1100, ₹1250, ... The amount at the end of the nth month is $500+150n$ (for example, $n=2$ gives $500+300=800$ ✓).

₹800, ₹950, ₹1100, ... ; amount in the nth month = ₹(500 + 150n).

The number decreases by 9 every hour, so after n hours there are $120-9n$ members.

111, 102, 93, ... members; at the end of the nth hour: 120 − 9n.

Area = length × breadth $=13b$.

$$\text{(i) }13\times 12=156{\text{cm}}^2\text{(ii) }13\times 10=130{\text{cm}}^2\text{(iii) }13\times 8=104{\text{cm}}^2$$

Each time the breadth decreases by 2 cm, the area decreases by $13\times 2=26{\text{cm}}^2$: 156, 130, 104, ... The pattern is $A=13b$, a linear expression in b.

(i) 156 cm² (ii) 130 cm² (iii) 104 cm²; linear pattern: A = 13b.

Volume = length × breadth × height $=7\times 11\times h=77h$.

$$\text{(i) }77\times 5=385{\text{cm}}^3\text{(ii) }77\times 9=693{\text{cm}}^3\text{(iii) }77\times 13=1001{\text{cm}}^3$$

The height increases by 4 cm each time and the volume increases by $77\times 4=308{\text{cm}}^3$ each time. The pattern is $V=77h$.

(i) 385 cm³ (ii) 693 cm³ (iii) 1001 cm³; linear pattern: V = 77h.

Pages left after n days $=500-20n$: 480, 460, 440, ... After 15 days:

$$500-20\times 15=500-300=200$$

200 pages will be left. Linear pattern: pages left after n days = 500 − 20n.

(i) Height after 7 months $=1.75+7\times 0.5=1.75+3.5=5.25$ feet.

(ii)

(iii) $h=1.75+0.5t$. This is linear growth because in every month (equal interval) the height increases by the same fixed amount, 0.5 feet.

(i) 5.25 feet (iii) h = 1.75 + 0.5t; linear growth because h increases by a constant 0.5 ft every month.

(i) Value after 3 years $=10000-3\times 800=10000-2400=₹7600$.

(ii)

(iii) $v=10000-800t$. This is linear decay because every year the value falls by the same fixed amount, ₹800.

(i) ₹7600 (iii) v = 10000 − 800t; linear decay because v decreases by a constant ₹800 every year.

(i) Population after 6 years $=750+6\times 50=750+300=1050$.

(ii)

(iii) $P=750+50t$. It is linear growth because the population increases by the same fixed number, 50, every year.

(i) 1050 (iii) P = 750 + 50t; linear growth because P increases by a constant 50 every year.

(i) $b(x)=600-15x$. It represents linear decay because the balance decreases by the same fixed amount, ₹15, every day.

(ii) The balance runs out when $b(x)=0$:

$$600-15x=0\Rightarrow 15x=600\Rightarrow x=40$$

(iii)

(i) b(x) = 600 − 15x (ii) The balance runs out after 40 days.

Substituting the two observations in $y=ax+b$:

$$400=10a+b\text{and}500=14a+b$$

From the first, $b=400-10a$. Substituting in the second:

$$500=14a+400-10a\Rightarrow 4a=100\Rightarrow a=25,b=400-250=150$$

So $y=25x+150$: a fixed fee of ₹150 and ₹25 per module. Check: $14\times 25+150=500$ ✓

a = 25, b = 150, so y = 25x + 150.

$$800=10a+b\text{and}1100=15a+b$$

Subtracting the first equation from the second: $300=5a$, so $a=60$. Then $b=800-10\times 60=200$.

So $y=60x+200$: a fixed fee of ₹200 and ₹60 per hour. Check: $15\times 60+200=1100$ ✓

a = 60, b = 200, so y = 60x + 200.

Write C for the Celsius reading and F for the Fahrenheit reading, so $C=aF+b$.

$$0=32a+b\text{and}100=212a+b$$

Subtracting: $100=180a$, so $a=\frac{100}{180}=\frac{5}{9}$. Then $b=-32a=-\frac{160}{9}$.

$$C=\frac{5}{9}F-\frac{160}{9}=\frac{5}{9}(F-32)$$

Check: $F=212$ gives $\frac{5}{9}\times 180=100$ ✓

$a=\frac{5}{9}$, $b=-\frac{160}{9}$, i.e. $C=\frac{5}{9}(F-32)$.

Each line passes through (0, 0). Second points: $y=4x$ through (1, 4); $y=2x$ through (2, 4); $y=x$ through (4, 4).

Here b = 0 in each, so every line passes through the origin. The value of a is the slope: the bigger a is, the steeper the line. All three rise from left to right because a > 0.

All three pass through the origin (b = 0); y = 4x is the steepest and y = x the least steep. Larger a means a steeper line.

Points: $y=-6x$ through (−1, 6) and (1, −6); $y=-3x$ through (−1, 3) and (1, −3); $y=-x$ through (−4, 4) and (4, −4).

Again b = 0, so all pass through the origin. Since a < 0, each line falls from left to right. The larger the size of a (6 > 3 > 1), the steeper the line.

All pass through the origin and slope downwards (a < 0); y = −6x is the steepest, y = −x the least steep.

Points: $y=5x$ through (1, 5) and (−1, −5); $y=-5x$ through (−1, 5) and (1, −5).

Both pass through the origin (b = 0) and are equally steep, but one rises and the other falls. Changing the sign of a reflects the line in the y-axis: the two lines are mirror images of each other.

The lines are equally steep, pass through the origin, and are mirror images in the y-axis (and in the x-axis).

Points: $y=3x-1$ through (0, −1) and (1, 2); $y=3x$ through (0, 0) and (1, 3); $y=3x+1$ through (0, 1) and (1, 4).

The value a = 3 is the same, so all three lines have the same steepness and are parallel. Only b changes: it is where the line cuts the y-axis, at (0, −1), (0, 0) and (0, 1). Changing b shifts the line up or down without turning it.

The lines are parallel (same a = 3); they cut the y-axis at −1, 0 and 1 (the values of b).

Points: $y=-2x-3$ through (0, −3) and (−2, 1); $y=-2x$ through (0, 0) and (2, −4); $y=2x+3$ through (0, 3) and (−2, −1).

$y=-2x-3$ and $y=-2x$ have the same a = −2, so they are parallel, and b shifts one 3 units below the other. $y=2x+3$ has a = +2: it is just as steep but rises instead of falling, so it cuts both other lines. (It meets $y=-2x-3$ at (−1.5, 0) and $y=-2x$ at (−0.75, 1.5).)

y = −2x − 3 ∥ y = −2x (same a); y = 2x + 3 has the opposite slope and crosses both. b gives the y-intercepts −3, 0 and 3.

We need the highest power to be 3 and the $x^2$ term to be $-7x^2$. The other coefficients can be chosen freely (the $x^3$ coefficient must not be 0).

For example, $x^3-7x^2+2x+5$.

$$\text{(i) }5(1{)}^2-3(1)+7=5-3+7=9$$

$$\text{(ii) }4a^3-a^2+6$$

(i) 9 (ii) $4a^3-a^2+6$

Let the number be $x$.

$$\frac{5}{2}x+\frac{2}{3}=-\frac{7}{12}$$

$$\frac{5}{2}x=-\frac{7}{12}-\frac{8}{12}=-\frac{15}{12}=-\frac{5}{4}$$

$$x=-\frac{5}{4}\times \frac{2}{5}=-\frac{1}{2}$$

Check: $\frac{5}{2}\times \left(-\frac{1}{2}\right)+\frac{2}{3}=-\frac{5}{4}+\frac{2}{3}=\frac{-15+8}{12}=-\frac{7}{12}$ ✓

The number is $-\frac{1}{2}$.

Let the smaller number be $x$; the other is $5x$. After adding 21 they become $x+21$ and $5x+21$. The larger new number is twice the smaller:

$$5x+21=2(x+21)\Rightarrow 5x+21=2x+42\Rightarrow 3x=21\Rightarrow x=7$$

The numbers are 7 and 35. Check: $7+21=28$ and $35+21=56=2\times 28$ ✓

(The other possibility, $x+21=2(5x+21)$, gives $x=-\frac{7}{3}$, which would make $5x$ negative, so it is rejected.)

The numbers are 7 and 35.

Amount after n months $=800+250n$: ₹1050, ₹1300, ₹1550, ...

$$\text{(i) }800+250\times 6=800+1500=₹2300$$

$$\text{(ii) 2 years}=24\text{ months: }800+250\times 24=800+6000=₹6800$$

(i) ₹2300 (ii) ₹6800; linear pattern: amount after n months = ₹(800 + 250n).

Let the tens digit be $a$ and the units digit be $b$. The number is $10a+b$ and the reversed number is $10b+a$.

$$(10a+b)+(10b+a)=143\Rightarrow 11(a+b)=143\Rightarrow a+b=13$$

The digits differ by 3, so $a-b=3$ or $b-a=3$.

• If $a-b=3$: adding to $a+b=13$ gives $2a=16$, so $a=8$, $b=5$. Number = 85.
• If $b-a=3$: $a=5$, $b=8$. Number = 58.

Either way the two numbers are 85 and 58. Check: $85+58=143$ ✓

The two numbers are 85 and 58.

First write each equation in the form $y=ax+b$; then a is the slope and b is the y-intercept, and the line cuts the y-axis at (0, b).

Points used for plotting: (i) (0, 4), (2, −2); (ii) (0, 3.5), (−1, 1.5); (iii) (0, −2), (5, 4); (iv) (0, −3.67), (4, 4.33).

Slopes: −3, 2, 6/5, 2; y-intercepts: 4, 7/2, −2, −11/3. Lines (ii) and (iv) are parallel, since both have slope 2.

(i) Put x = 313:

$$y=\frac{9}{5}(313-273)+32=\frac{9}{5}\times 40+32=72+32=104$$

(ii) Put y = 158:

$$158=\frac{9}{5}(x-273)+32\Rightarrow 126=\frac{9}{5}(x-273)\Rightarrow x-273=126\times \frac{5}{9}=70\Rightarrow x=343$$

(i) 104 °F (ii) 343 K

Work = force × distance. With force = 3 units:

$$w=3d$$

Table of values: d = 0, 1, 2, 3 gives w = 0, 3, 6, 9. Plot d along the horizontal axis and w along the vertical axis. Distance cannot be negative, so the graph starts at the origin.

When d = 2, $w=3\times 2=6$. On the graph, go up from d = 2 to the line and across to the w-axis: it reads 6 ✓

w = 3d; when d = 2 units, the work done is 6 units (the point (2, 6) lies on the graph).

(i) Let $p(x)=ax+b$. Then $p(1)=5$ and $p(3)=11$:

$$a+b=5,3a+b=11$$

Subtracting: $2a=6$, so $a=3$ and $b=2$. Hence $p(x)=3x+2$.

(ii) y-axis (x = 0): $p(0)=2$, so the point is (0, 2). x-axis ($p(x)=0$): $3x+2=0$, so $x=-\frac{2}{3}$; the point is $\left(-\frac{2}{3},0\right)$.

(iii)

The line through (1, 5) and (3, 11) cuts the y-axis at 2 and the x-axis a little to the left of the origin, at about −0.67 ✓

(i) p(x) = 3x + 2 (ii) It cuts the y-axis at (0, 2) and the x-axis at (−2/3, 0).

From (i): $p(0)=b=5$.

From (iii): $(a+c)x+(b+d)=6x+4$ for all x, so $a+c=6$ and $b+d=4$. Hence $d=4-5=-1$.

From (ii): $p(3)-q(3)=0$, i.e. $(3a+5)-(3c-1)=0$, so $3(a-c)=-6$ and $a-c=-2$.

Solving $a+c=6$ and $a-c=-2$: $2a=4$, so $a=2$ and $c=4$.

Check: $p(x)+q(x)=(2x+5)+(4x-1)=6x+4$ ✓; $p(x)-q(x)=-2x+6=0$ at $x=3$ ✓

p(x) = 2x + 5 and q(x) = 4x − 1.

(i) Stage 1 is one hexagon: 6 matchsticks. Each new hexagon shares one side with the previous one, so it needs only 5 new matchsticks.

Stage 4 needs $6+5+5+5=21$ matchsticks and Stage 5 needs $21+5=26$.

(ii)

(iii) The number increases by 5 each time, starting with 6 = 5 × 1 + 1. So the nth stage needs $5n+1$ matchsticks.

(iv) $5\times 15+1=76$ matchsticks.

(v) $5n+1=200$ gives $5n=199$, so $n=39.8$, which is not a whole number. So 200 matchsticks cannot form a stage exactly. (Stage 39 uses 196 and stage 40 uses 201.)

(i) 21 and 26 (iii) 5n + 1 (iv) 76 (v) No, since 5n + 1 = 200 gives n = 39.8, not a whole number.

From (i): $2a+b=3$ and $6a+b=11$. Subtracting: $4a=8$, so $a=2$ and $b=3-4=-1$. So $p(x)=2x-1$.

From (iii): parallel lines have the same slope, so $c=2$. From (ii): $q(4)=2\times 4+d=-1$, so $d=-9$. So $q(x)=2x-9$.

x-axis points: $2x-1=0$ gives $x=\frac{1}{2}$; $2x-9=0$ gives $x=\frac{9}{2}$.

p(x) = 2x − 1 meets the x-axis at (1/2, 0); q(x) = 2x − 9 meets the x-axis at (9/2, 0).

Factorise: $f(x)=a(x+1)$. Then:

• $f(-1)=a(0)=0$ for every a, so every such line passes through the point (−1, 0) on the x-axis.
• The slope and the y-intercept are equal (both are a), so each line cuts the y-axis at (0, a).
• Since a > 0, every such line rises from left to right (they all show linear growth).

All of them pass through (−1, 0), have slope equal to their y-intercept (a), and rise from left to right.