What is the multiplication identity used throughout this chapter?

For any numbers x, a, b: (x+a)(x+b) = x² + (a+b)x + ab. This single identity covers every sign combination — since a and b can be negative, it also handles products like (x−a)(x+b) or (x−a)(x−b) without needing separate rules.

How do you factor a quadratic like x²+px+q?

Find two numbers a and b whose sum is p and whose product is q, then write x²+px+q = (x+a)(x+b). If the sum or product is negative, at least one of a, b must be negative.

What do you do when the factor pair isn't easy to guess?

Use the identity (a−b)² = (a+b)² − 4ab to compute the difference a−b algebraically from the known sum and product. Combining a+b and a−b then gives a and b directly, without trial and error.

How does factoring a polynomial help solve a quadratic equation?

Once a quadratic is factored as (x+a)(x+b) = 0, the equation is solved immediately: since a product of two numbers is zero only when at least one of them is zero, x = −a or x = −b.

Where does the quadratic formula come from?

Applying the same sum-and-product factoring method to the general equation ax²+bx+c=0 (after dividing by a) derives x = (−b ± √(b²−4ac)) / 2a — the quadratic formula is the general-purpose version of this chapter's factoring technique, not a separate rule.

Chapter 9 · Kerala SSLC Class 10 Maths

Polynomials and Equations

Two ways to learn — a crisp formula reference, or a story that builds from a mental-math trick to the general quadratic formula.

(x+a)(x+b) identitySum & product factoringSum-difference trickQuadratic formula

Also useful:Chapter 5 — Second Degree Equations Practice Problems — Ch 9 (solved questions)1-on-1 Kerala SSLC Maths Tuition

Kerala SSLCClass 10MathematicsChapter 9⏱ 12 min crisp · 18 min story

Five scenes from a mental-math trick to the general quadratic formula.

How this works

Five scenes build from a mental-math multiplication trick all the way to the general quadratic formula. Read the story, then tap "The Maths" for the formal result.

The Mental Math Trick

Can you multiply 38 × 32 in your head? Try writing 38 as 30+8 and 32 as 30+2.

38 × 32 = (30+8)(30+2)

= (30×30) + (30×2) + (8×30) + (8×2)

= 900 + 300 + 16

= 1216

The middle two products combine neatly: (30×2)+(8×30) = (2+8)×30 = 10×30 = 300. Try the same trick on 48×42 and 78×72.

48×42 = 40² + ((8+2)×40) + (8×2) = 1600+400+16 = 2016

78×72 = 4900+700+16 = 5616

Each of these fits a single pattern: (x+2)(x+8) = x²+10x+16, where x stands for the shared tens part (30, 40, 70).

The general rule

For ANY numbers x, a, b: (x+a)(x+b) = x² + (a+b)x + ab.

This holds even when a or b is negative — since x−1 is really x+(−1), the same identity handles subtraction too.

(x+4)(x−1) = (x+4)(x+(−1))

= x² + (4+(−1))x + (4×(−1))

= x² + 3x − 4

Working Backwards

Now flip the question: can x²+5x+6 be written as a product of two first-degree polynomials?

Matching x²+(a+b)x+ab with x²+5x+6 means finding a, b with a+b=5 and ab=6. A little thought gives 2 and 3.

x²+5x+6 = (x+2)(x+3)

A twist

What about x²−5x+6? Same product (6), but now the sum must be −5.

Write 6 as a product of two NEGATIVE numbers: 6 = (−2)×(−3), and (−2)+(−3) = −5. That's the pair.

x²−5x+6 = (x+(−2))(x+(−3)) = (x−2)(x−3)

One more twist

And x²+5x−6? Now the PRODUCT is negative, so one of a, b must be negative and the other positive.

Trying −6=(−1)×6 gives sum 5. That's the pair.

x²+5x−6 = (x+6)(x−1)

In every case — positive/positive, negative/negative, or mixed signs — the same identity (x+a)(x+b)=x²+(a+b)x+ab does the job. Using signed numbers means we never need a separate rule for each sign combination.

When Guessing Fails

Try factoring x²+27x+180. What two numbers have sum 27 and product 180?

Listing factor pairs of 180 and checking each sum by hand would take a while. There is a faster, algebraic way.

Recall from Class 8

The square of a difference can be found from the square of a sum and the product: (a−b)² = (a+b)² − 4ab.

(a−b)² = 27² − 4(180)

= 729 − 720

= 9

a−b = 3 or a−b = −3

Combine a+b=27 with a−b=3 (or its opposite) to get a and b directly — no guessing needed.

a = (27+3)/2 = 15

b = (27−3)/2 = 12

So x²+27x+180 = (x+15)(x+12). (Taking a−b=−3 instead just swaps which number is called a and which is b — the factorization is identical.)

From Factors to Solutions

Here's why factoring matters beyond simplifying expressions. Take the equation x²−4x+3=0. Factor the left side first.

x²−4x+3 = (x−1)(x−3)

So the equation becomes: (x−1)(x−3) = 0

For a product of two numbers to be zero, at least one of them must BE zero.

x−1=0 or x−3=0

x=1 or x=3

This gives the same two solutions as completing the square did in the earlier chapter — but reaching them straight from the factored form.

A real problem

One side of a rectangle is 3 metres longer than the other, and its area is 270 square metres. Find the sides.

Let the shorter side be x. Then x(x+3)=270, so x²+3x−270=0. Factoring needs a+b=3, ab=−270 — not an easy guess, so use the sum-difference trick.

(a−b)² = 3² − 4(−270) = 9+1080 = 1089 = 33²

a−b = 33 → a=(3+33)/2=18, b=(3−33)/2=−15

x²+3x−270 = (x−15)(x+18)

(x−15)(x+18)=0 → x=15 or x=−18

Since x is a rectangle's side, it can't be negative — so x=−18 is rejected. The shorter side is 15 m, the longer side 18 m.

The Formula for Every Quadratic

Everything so far assumed the x² coefficient was exactly 1. What about 2x²+3x−2=0?

Factor out 2 first, so the bracket has a leading coefficient of 1: 2x²+3x−2 = 2[x²+(3/2)x−1].

Need a+b=3/2, ab=−1

(a−b)² = (3/2)² − 4(−1) = 9/4 + 4 = 25/4

a−b = 5/2

a=(3/2+5/2)/2=2, b=(3/2−5/2)/2=−1/2

So 2x²+3x−2 = 2(x+2)(x−1/2), giving solutions x=−2 or x=1/2.

Doing this once and for all

What if we run the SAME method on the general equation ax²+bx+c=0, without picking specific numbers?

Divide through by a: x²+(b/a)x+(c/a)=0. Factor the bracket with sum p+q=−b/a and product pq=c/a (matching the (x+p)(x+q)=0 form used to solve).

(p−q)² = (p+q)² − 4pq = (b/a)² − 4(c/a) = (b²−4ac)/a²

p−q = √(b²−4ac) / a

p = [−b + √(b²−4ac)] / 2a

q = [−b − √(b²−4ac)] / 2a

The solutions x=−p or x=−q become:

x = [−b ± √(b²−4ac)] / 2a

This is the quadratic formula — not a separate rule, but the same sum-and-product factoring method, carried out symbolically instead of with specific numbers.

The Big Picture

Every idea in this chapter is the same identity, used forwards, backwards, and finally in general.

Expand (x+a)(x+b)?

x² + (a+b)x + ab

Factor x²+px+q?

find a,b with a+b=p, ab=q

Sum/product hard to guess?

(a−b)² = (a+b)² − 4ab

Solve a factored equation?

(x+a)(x+b)=0 → x=−a or −b

Leading coefficient isn't 1?

divide out a, factor, multiply back

Need a formula for ANY quadratic?

x = (−b ± √(b²−4ac)) / 2a

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