Introduction to Linear Polynomials

The degree of a polynomial in one variable is the highest power of the variable that appears in it.

(i) \(2x^{2}-5x+3\) — the highest power of \(x\) is 2, so the degree is 2.

(ii) \(y^{3}+2y-1\) — the highest power of \(y\) is 3, so the degree is 3.

(iii) \(-9\) — a non-zero constant can be written as \(-9x^{0}\), so the degree is 0.

(iv) \(4z-3\) — the highest power of \(z\) is 1, so the degree is 1. This is a linear polynomial.

Any polynomial whose highest power matches the required degree is acceptable. Here is one example of each.

Degree 1 (linear):

Degree 2 (quadratic):

Degree 3 (cubic):

The only condition is that the coefficient of the highest-power term must not be zero — otherwise that term vanishes and the degree drops.

The coefficient of a term is the number multiplying that power of the variable, including its sign.

Looking at the polynomial term by term:

The \(x^{2}\) term is \(+6x^{2}\), so the coefficient of \(x^{2}\) is 6.

The \(x^{3}\) term is \(-3x^{3}\), so the coefficient of \(x^{3}\) is −3.

Common mistake: writing 3 instead of −3. The minus sign belongs to the coefficient.

The polynomial has a \(z^{3}\) term, a \(z^{2}\) term and a constant term, but no \(z\) term is written.

A missing term is the same as a term with coefficient zero. We may write the polynomial as:

So the coefficient of \(z\) is 0.

The constant term is the term that contains no variable — the value the polynomial takes when \(x=0\).

Here that term is \(-10\), so the constant term is −10.

Check by substituting \(x=0\):

Substitute each value of \(x\) into \(5x-3\).

(i) \(x=0\):

(ii) \(x=-1\):

(iii) \(x=2\):

So the values are −3, −8 and 7.

(i) \(s=0\):

(ii) \(s=-3\):

(iii) \(s=4\):

So the values are 6, 81 and 102.

Careful: in part (ii), \((-3)^{2}=+9\), and \(-4\times(-3)=+12\). Both signs turn positive.

Setting up. Let Salil’s present age be \(x\) years. Then his mother’s present age is \(3x\) years.

After 5 years, Salil will be \(x+5\) and his mother will be \(3x+5\). Their sum is 70:

Answer: Salil is 15 years old and his mother is \(3\times15=\) 45 years old.

Check: after 5 years they will be 20 and 50, and \(20+50=70\). ✓

Setting up. Since the ratio is \(2:5\), let the integers be \(2x\) and \(5x\). The larger minus the smaller is 63:

Answer: the integers are \(2\times21=\) 42 and \(5\times21=\) 105.

Check: \(105-42=63\) ✓ and \(42:105=2:5\) ✓

Setting up. Let the number of five-rupee coins be \(x\). Then the number of two-rupee coins is \(3x\).

Value of the two-rupee coins is \(2\times3x=6x\), and value of the five-rupee coins is \(5x\). The total is ₹88:

Answer: Ruby has 8 five-rupee coins and \(3\times8=\) 24 two-rupee coins.

Check: \(24\times2+8\times5=48+40=88\) ✓

Watch out: the equation is about value in rupees, not about the number of coins. Multiply each count by the value of that coin.

Let the shorter piece be \(x\) feet. Then the longer piece is \(4x\) feet, and together they make 300 feet:

Answer: the pieces are 60 feet and \(4\times60=\) 240 feet.

Check: \(60+240=300\) ✓

Let the width be \(x\) cm. Then the length is \(2x+3\) cm.

The perimeter of a rectangle is \(2(\text{length}+\text{width})\):

Answer: width \(=\) 3 cm, length \(=2(3)+3=\) 9 cm.

Check: perimeter \(=2(9+3)=24\) cm ✓

She starts with ₹500 and adds ₹150 each month.

From the second month onwards the amounts are ₹800, ₹950, ₹1100, …

Linear expression for the \(n\)th month:

The amount increases by the constant value ₹150 each month, so this is a linear pattern.

Start with 120 members and subtract 9 for each hour that passes.

Linear expression:

Because a constant number is subtracted each hour, this is a linear pattern — an example of linear decay.

Area of a rectangle \(=\) length \(\times\) breadth, and the length is fixed at 13 cm.

(i) \(13\times12=\) 156 cm²

(ii) \(13\times10=\) 130 cm²

(iii) \(13\times8=\) 104 cm²

Linear pattern. If the breadth is \(b\) cm, then:

This is a linear polynomial in \(b\). Every increase of 1 cm in breadth raises the area by exactly 13 cm².

Volume \(=\) length \(\times\) breadth \(\times\) height \(=7\times11\times h=77h\).

(i) \(77\times5=\) 385 cm³

(ii) \(77\times9=\) 693 cm³

(iii) \(77\times13=\) 1001 cm³

Linear pattern:

Although volume is a three-dimensional quantity, two of the three dimensions are fixed here, so the volume is a linear function of the height alone.

After \(d\) days she has read \(20d\) pages, so the number of pages left is:

After 15 days:

200 pages will be left.

This is a linear pattern showing linear decay — a constant 20 pages is removed each day. The book is finished when \(500-20d=0\), that is on day \(d=25\).

(iii) The expression first — starting height 1.75 ft, increasing 0.5 ft per month:

(i) Height after 7 months:

The height after 7 months is 5.25 feet.

(ii) Table of values

Why it is linear growth: \(h(t)\) is a polynomial of degree 1 in \(t\), and the height increases by the same amount, 0.5 ft, for every one-month increase. A constant increase over equal intervals is exactly what linear growth means.

(iii) The expression:

(i) Value after 3 years:

The value after 3 years is ₹7,600.

(ii) Table of values

Why it is linear decay: the value falls by the same amount, ₹800, every year. A constant decrease over equal intervals is linear decay. This kind of fall in value is called straight-line depreciation.

(iii) The expression:

(i) Population after 6 years:

The population after 6 years is 1050.

(ii) Table of values

Why it is linear growth: exactly 50 people are added each year, a constant increase over equal intervals, and \(P(t)\) is a linear polynomial in \(t\).

(i) The model. The balance starts at ₹600 and ₹15 is deducted daily:

This represents linear decay because the same amount, ₹15, is subtracted for each additional day — a constant decrease over equal intervals.

(ii) When the balance runs out the remaining balance is zero:

The balance will run out after 40 days.

(iii) Table of values

Substitute the two observations into \(y=ax+b\):

Subtract the first from the second to eliminate \(b\):

Substitute back:

Answer: \(a=\) 25 and \(b=\) 150, so the bill is \(y=25x+150\).

What the numbers mean: \(a=25\) is the cost of one module and \(b=150\) is the fixed monthly fee.

Substituting the two observations:

Subtracting:

Substituting back:

Answer: \(a=\) 60 and \(b=\) 200, so \(y=60x+200\).

The court costs ₹60 per hour and the fixed monthly fee is ₹200.

Substitute the two fixed points into \({}^\circ C=a\,{}^\circ F+b\).

Ice melts — when \({}^\circ F=32\), \({}^\circ C=0\):

Water boils — when \({}^\circ F=212\), \({}^\circ C=100\):

Subtracting the first from the second:

Substituting back:

Answer: \(a=\dfrac{5}{9}\) and \(b=-\dfrac{160}{9}\), giving the familiar conversion formula:

Check: at \({}^\circ F=212\), \(\frac{5}{9}(212-32)=\frac{5}{9}(180)=100\) ✓

Draw each set on the same pair of axes. Plot two points per line — usually \(x=0\) and \(x=1\) — and join them with a ruler.

(i) \(y=4x,\;y=2x,\;y=x\). All three pass through the origin because \(b=0\). As \(a\) grows from 1 to 4 the line becomes steeper. So \(a\) controls the steepness, or slope.

(ii) \(y=-6x,\;y=-3x,\;y=-x\). Again all pass through the origin, but now they slope downwards from left to right because \(a\) is negative. The larger the size of \(a\), the steeper the fall.

(iii) \(y=5x\) and \(y=-5x\). Equal steepness but opposite directions — the two lines are mirror images of each other in the \(y\)-axis.

(iv) \(y=3x-1,\;y=3x,\;y=3x+1\). All three have the same \(a=3\), so they are parallel. Changing \(b\) slides the line up or down; each line cuts the \(y\)-axis at \(y=b\), that is at \(-1\), \(0\) and \(1\).

(v) \(y=-2x-3,\;y=-2x,\;y=2x+3\). The first two are parallel (both have \(a=-2\)); the third has \(a=+2\) and therefore slopes the other way and is not parallel to them.

Conclusion. In \(y=ax+b\), the coefficient \(a\) is the slope — it fixes the steepness and the direction — and the constant \(b\) is the \(y\)-intercept, the point where the line crosses the \(y\)-axis. Two lines are parallel exactly when they have the same \(a\).

Two conditions must hold: the highest power of \(x\) must be 3, and the number multiplying \(x^{2}\) must be \(-7\).

One correct answer is:

The coefficient of \(x^{3}\) may be any non-zero number, and the \(x\) term and constant term may be anything at all. Another equally correct answer is \(2x^{3}-7x^{2}+1\).

(i) Substituting \(x=1\):

The value is 9.

(ii) Substituting \(t=a\):

Here the answer stays in symbolic form. Substituting one letter for another simply renames the variable — the value is \(4a^{3}-a^{2}+6\).

Let the number be \(x\). Translating the sentence into an equation:

Move the constant to the right:

Writing both fractions with denominator 12:

Multiply both sides by \(\frac{2}{5}\):

Answer: the number is \(-\dfrac{1}{2}\).

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

Let the smaller number be \(x\). Then the other number is \(5x\).

After adding 21 to each, the numbers become \(x+21\) and \(5x+21\). The larger new number is twice the smaller new number:

Answer: the numbers are 7 and \(5\times7=\) 35.

Check: adding 21 gives 28 and 56, and \(56=2\times28\) ✓

The linear pattern. Let \(m\) be the number of months:

(i) After 6 months:

You will have ₹2,300.

(ii) After 2 years — that is 24 months:

You will have ₹6,800.

Careful: the rate is per month, so 2 years must be converted to 24 months before substituting.

Let the tens digit be \(a\) and the units digit be \(b\). Then the number is \(10a+b\), and the number with digits interchanged is \(10b+a\).

Their sum is 143:

The digits differ by 3:

Case 1: \(a-b=3\). Adding to \(a+b=13\) gives \(2a=16\), so \(a=8\) and \(b=5\). The number is 85.

Case 2: \(a-b=-3\). Then \(2a=10\), so \(a=5\) and \(b=8\). The number is 58.

Answer: the two numbers are 58 and 85.

Check: \(58+85=143\) ✓, and the digits 5 and 8 differ by 3 ✓. Notice that the two numbers are each other’s reverse, which is why both satisfy the conditions.

First rewrite each equation in the form \(y=ax+b\), where \(a\) is the slope and \(b\) the \(y\)-intercept.

Are any parallel? Yes. Lines (ii) and (iv) both have slope 2, so they are parallel. Their \(y\)-intercepts differ, so they are distinct parallel lines and never meet.

To draw each line, plot the \(y\)-intercept from the last column and one more point — for (i) take \(x=1\) to get \((1,1)\) — then join with a ruler.

(i) Substituting \(x=313\):

The temperature is 232 °F.

(ii) Substituting \(y=158\) and solving for \(x\):

The temperature is 298.2 K, which may also be written as \(\dfrac{1491}{5}\) K.

The general relation. If the constant force is \(F\), then work \(=\) force \(\times\) distance:

Taking the force as 3 units:

This is a linear equation in two variables. Because there is no constant term, its graph is a straight line through the origin.

Work done when \(d=2\):

The work done is 6 units.

Table for the graph

Plot these points, join them with a ruler, then read the graph at \(d=2\). You should land exactly on \(w=6\), which verifies the calculation.

(i) A linear polynomial has the form \(p(x)=ax+b\). Substituting the two given points:

Subtracting the first equation from the second:

Substituting back, \(3+b=5\), so \(b=2\). Therefore:

(ii) Where the graph cuts the axes.

It cuts the \(y\)-axis where \(x=0\), giving \(p(0)=2\) — the point \((0,2)\).

It cuts the \(x\)-axis where \(p(x)=0\):

— the point \(\left(-\frac{2}{3},\,0\right)\).

(iii) Verification. Plot \((1,5)\) and \((3,11)\) and join them. The line should pass through \((0,2)\) on the \(y\)-axis and cross the \(x\)-axis a little to the left of the origin, at about \(-0.67\). ✓

Take the three conditions one at a time.

Condition (i): \(p(0)=5\). Since \(p(0)=b\), we get:

Condition (iii): \(p(x)+q(x)=6x+4\). Adding the two polynomials:

Comparing coefficients on both sides:

Since \(b=5\), we get \(d=4-5=-1\).

Condition (ii): \(p(x)-q(x)\) cuts the \(x\)-axis at \((3,0)\), which means \(p(3)-q(3)=0\):

Solving the pair \(a+c=6\) and \(a-c=-2\): adding gives \(2a=4\), so \(a=2\) and \(c=4\).

Answer:

Check: \(p(0)=5\) ✓. \(p(x)+q(x)=6x+4\) ✓. \(p(x)-q(x)=-2x+6\), which is zero at \(x=3\) ✓

Understanding the pattern. The first hexagon needs 6 matchsticks. Every hexagon after that shares one side with the previous one, so it needs only 5 new matchsticks rather than 6.

(i) and (ii) The table. Stage 4 needs \(16+5=21\) and stage 5 needs \(21+5=26\) matchsticks.

(iii) The rule. The count goes up by a constant 5 each stage, so it is a linear pattern of the form \(5n+k\). At \(n=1\) the value is 6, so \(5+k=6\) and \(k=1\):

This makes sense: think of one matchstick standing alone at the start, then 5 added for every hexagon.

(iv) The 15th stage:

76 matchsticks are required.

(v) Can 200 matchsticks form a stage? Set \(M(n)=200\):

No. A stage number must be a whole number, and \(39.8\) is not. Another way to see it: every value of \(5n+1\) ends in 6 or 1, and 200 ends in 0. Stage 39 uses 196 matchsticks and stage 40 uses 201, so 200 is skipped.

Finding \(p(x)\). Substituting the two points into \(p(x)=ax+b\):

Subtracting the first from the second:

Then \(2(2)+b=3\), so \(b=-1\), giving:

Finding \(q(x)\). Parallel lines have equal slopes, so \(c=a=2\). Using the point \((4,-1)\):

Where they meet the \(x\)-axis — set each polynomial equal to zero.

For \(p(x)\): \(2x-1=0\) gives \(x=\frac{1}{2}\), the point \(\left(\frac{1}{2},\,0\right)\).

For \(q(x)\): \(2x-9=0\) gives \(x=\frac{9}{2}\), the point \(\left(\frac{9}{2},\,0\right)\).

The two lines are parallel, so they never meet each other — they only meet the \(x\)-axis, at two different points.

Take out the common factor \(a\):

They all pass through the same point. Whatever the value of \(a\), substituting \(x=-1\) gives:

So every such line passes through \((-1,\,0)\). They all cut the \(x\)-axis at exactly the same point, \(x=-1\).

Other shared features. Since \(a>0\), every one of these lines has a positive slope, so all of them rise from left to right. Each cuts the \(y\)-axis at \((0,a)\), a different point for each value of \(a\).

Picture it: the family \(f(x)=a(x+1)\) is a fan of lines pivoting about the fixed point \((-1,0)\), with \(a\) deciding how steeply each one rises.