Polynomials – Class IX [CBSE]

You have studied algebraic expressions, their addition, subtraction, multiplication and division in earlier classes. You also have studied how to factorize some algebraic expressions. You may recall the algebraic identities:

(x + y)2 = x2 + 2xy + y2

(x – y)2 = x2 – 2xy + y2

and x2– y2 = (x + y) (x – y)

and their use in factorisation. In this chapter, we shall start our study with a particular type of algebraic expression, called polynomialpolynomialpolynomial, and the terminology related to it. We shall also study the Remainder Theorem and Factor Theorem and their use in the factorisation of polynomials. In addition to the above, we shall study some more algebraic identities and their use in factorisation and in evaluating some given expressions.

Polynomials in One Variable:

There is a difference between a letter denoting a constant and a letter denoting a variable. The values of the constants remain the same throughout a particular situation, that is, the values of the constants do not change in a given problem, but the value of a variable can keep changing.

In the polynomial x2 + 2x, the expressions x2 and 2x are called the terms of the polynomial. Each term of a polynomial has a coefficient.

The degree of a non-zero constant polynomial is zero. The highest power of the variable in a polynomial as the degree of the polynomial.

Example 1 : Find the degree of each of the polynomials given below:

(i) x5 – x4 + 3

(ii) 2 – y2 – y3 + 2y8

(iii) 2

Solution :

(i) The highest power of the variable is 5. So, the degree of the polynomial is 5.

(ii) The highest power of the variable is 8. So, the degree of the polynomial is 8.

(iii)The only term here is 2 which can be written as 2x0 . So the exponent of x is 0. Therefore, the degree of the polynomial is 0.


Polynomial – Exercise 2.1


Zeroes of a Polynomial:

Consider the polynomial p(x) = 5x3– 2x2 + 3x – 2.

If we replace x by 1 everywhere in p(x), we get

p(1) = 5 × (1)3 – 2 × (1)2 + 3 × (1) – 2

= 5 – 2 + 3 –2

= 4

So, we say that the value of p(x) at x = 1 is 4.

Similarly, p(0) = 5(0)3 – 2(0)2 + 3(0) –2 = –2

 

Example 2: Find the value of each of the following polynomials at the indicated value of variables:

(i) p(x) = 5x2 – 3x + 7 at x = 1

(ii) q(y) = 3y3 – 4y + √11 at y = 2

(iii) p(t) = 4t4 + 5t3 – t2 + 6 at t = a

Solution:

(i) p(x) =  5x2 – 3x + 7

The value of the polynomial p(x) at x = 1  is given by

p(1) = 5(1)2 – 3(1) + 7 = 5 – 3 + 7 = 9

(ii) q(y) = 3y3 – 4y + √11

The value of the polynomial q(y) at x = 2  is given by

q(2) = 3(2)3 – 4(2) + √11  = 3×8 – 8 + √11   = 24 – 8 + √11  = 16 + √11

(iii) p(t) = 4t4 + 5t3 – t2 + 6 at t = a

The value of the polynomial p(t) at t = a  is given by

p(a = 4a4 + 5a3 – a2 + 6

 

Example 3: Check whether –2 and 2 are zeroes of the polynomial x + 2.

Solution : Let p(x) = x + 2.

Then p(2) = 2 + 2 = 4, p(–2) = –2 + 2 = 0

Therefore, –2 is a zero of the polynomial x + 2, but 2 is not.

 

Example 4 : Find a zero of the polynomial p(x) = 2x + 1.

Solution : Finding a zero of p(x), is the same as solving the equation p(x) = 0

Now, 2x + 1 = 0 gives us x = –1/2

So, –1/2 is a zero of the polynomial 2x + 1.

Now, if p(x) = ax + b, a ≠ 0, is a linear polynomial, how can we find a zero of p(x)? Example 4 may have given you some idea. Finding a zero of the polynomial p(x), amounts to solving the polynomial equation p(x) = 0.

Now, p(x) = 0 means ax + b = 0, a ≠ 0

So, ax = –b i.e., x = – b/a.

So, x = –b/a  is the only zero of p(x), i.e., a linear polynomial has one and only one zero.

Now we can say that 1 is the zero of x – 1, and –2 is the zero of x + 2.

 

Example 5: Verify whether 2 and 0 are zeroes of the polynomial x2 – 2x

Solution:

Letp(x) = x2 – 2x

Then p(2) = 22– 4 = 4 – 4 = 0 and p(0) = 0 – 0 = 0

Hence, 2 and 0 are both zeroes of the polynomial x2 – 2x.

 

Let us now list our observations:

(i) A zero of a polynomial need not be 0.

(ii) 0 may be a zero of a polynomial.

(iii) Every linear polynomial has one and only one zero.

(iv) A polynomial can have more than one zero.


Polynomials – Exercise 2.2


Remainder Theorem:

Let us consider two numbers 15 and 6. You know that when we divide 15 by 6, we get the quotient 2 and remainder 3. Do you remember how this fact is expressed? We write 15 as 15 = (2 × 6) + 3

We observe that the remainder 3 is less than the divisor 6. Similarly, if we divide 12 by 6, we get 12 = (2 × 6) + 0

What is the remainder here? Here the remainder is 0, and we say that 6 is a factor of 12 or 12 is a multiple of 6.

Now, the question is: can we divide one polynomial by another? To start with, let us try and do this when the divisor is a monomial. So, let us divide the polynomial 2x3 + x2 + x by the monomial x.

We have

(2x3 + x2 + x) ÷ x = 2x3/x + x2/x + x/x

= 2x2 + x + 1

In fact, you may have noticed that x is common to each term of 2x3 + x2 + x. So we can write 2x3+ x2 + x as x(2x2 + x + 1).

We say that x and 2x2 + x + 1 are factors of 2x3 + x2 + x, and 2x3 + x2 + x is a multiple of x as well as a multiple of 2x2 + x + 1.

Consider another pair of polynomials 3x2 + x + 1 and x.

Here, (3x2 + x + 1) ÷ x = (3x2 ÷ x) + (x ÷ x) + (1 ÷ x).

We see that we cannot divide 1 by x to get a polynomial term. So in this case we stop here, and note that 1 is the remainder. Therefore, we have

3x2 + x + 1 = {(3x + 1) × x} + 1

In this case, 3x + 1 is the quotient and 1 is the remainder. Do you think that x is a factor of 3x2 + x + 1? Since the remainder is not zero, it is not a factor.

Now let us consider an example to see how we can divide a polynomial by any non-zero polynomial.


Example 6: Divide p(x) by g(x), where p(x) = x + 3x2 – 1 and g(x) = 1 + x.

Solution : We carry out the process of division by means of the following steps:

Step 1 : We write the dividend x + 3x2 – 1 and the divisor 1 + x in the standard form, i.e., after arranging the terms in the descending order of their degrees. So, the dividend is 3x2 + x –1 and divisor is x + 1.

2.pngStep 2: We divide the first term of the dividend by the first term of the divisor, i.e., we divide 3x2 by x, and get 3x. This gives us the first term of the quotient.

3x2/x = 3x = first term

 

Step 3: We multiply the divisor by the first term of the quotient, and subtract this product from the dividend, i.e., we multiply x + 1 by 3x and subtract the product 3x2 + 3x from the dividend 3x2 + x – 1. This gives us the remainder as –2x – 1.

Step 4: We treat the remainder –2x – 1 as the new dividend. The divisor remains the same. We repeat Step 2 to get the next term of the quotient, i.e., we divide the first term – 2x of the (new) dividend by the first term x of the divisor and obtain – 2. Thus, – 2 is the second term in the quotient.

-2x/x = – 2 = second term of quotient

Step 5 : We multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is, we multiply x + 1 by – 2 and subtract the product – 2x – 2 from the dividend – 2x – 1. This gives us 1 as the remainder.

This process continues till the degree of the new dividend is less than the degree of the divisor. At this stage, this new dividend becomes the remainder and the sum of the quotients gives us the whole quotient.

Step 6 : Thus, the quotient in full is 3x – 2 and the remainder is 1.

Let us look at what we have done in the process above as a whole:

1

Notice that 3x2 + x – 1 = (x + 1) (3x – 2) + 1

i.e., Dividend = (Divisor × Quotient) + Remainder

 

In general, if p(x) and g(x) are two polynomials such that degree of p(x) ≥ degree of g(x) and g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that: p(x) = g(x)q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x). Here we say that p(x) divided by g(x), gives q(x) as quotient and r(x) as remainder.

In the example above, the divisor was a linear polynomial. In such a situation, let us see if there is any link between the remainder and certain values of the dividend.

Example 7: Divide the polynomial 3x4 – 4x3 – 3x – 1 by x – 1

Solution:

Divide the polynomial 3x4 – 4x3 – 3x – 1 by x – 1

Here the remainder is – 5. Now the zero of x – 1 is 1. So, putting x = 1 in p(x), we see that

p() = 3(1)4 – 4(1)3 – 3(1) – 1

= 3 – 4 – 3 – 1

= – 5 , which is the remainder.


Example 8: Find the remainder obtained  on dividing p(x) = x3 +  1 by x + 1

Solution:

By long division,

 

Find the remainder obtained  on dividing p(x) = x3 +  1 by x + 1

Here p(x) = x3 + 1, and the root of x + 1 = 0 is x = –1. We see that

p(–1) = (–1)3 + 1

= –1 + 1

= 0, which is equal to the remainder obtained by actual division.

Is it not a simple way to find the remainder obtained on dividing a polynomial by a linear polynomial? We shall now generalize this fact in the form of the following theorem. We shall also show you why the theorem is true, by giving you a proof of the theorem.


Remainder Theorem : Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).

Proof : Let p(x) be any polynomial with degree greater than or equal to 1. Suppose that when p(x) is divided by x – a, the quotient is q(x) and the remainder is r(x), i.e., p(x) = (x – a) q(x) + r(x)

Since the degree of x – a is 1 and the degree of r(x) is less than the degree of x – a, the degree of r(x) = 0. This means that r(x) is a constant, say r.

So, for every value of x, r(x) = r.

Therefore, p(x) = (x – a) q(x) + r

In particular, if x = a, this equation gives us p(a) = (a – a) q(a) + r = r, which proves the theorem.


Let us use this result in another example:

Example 9: Find the remainder when x4 + x3 – 2x2 + x + 1 is divided by x – 1.

Solution : Here, p(x) = x4 + x3 – 2x2 + x + 1, and the zero of x – 1 is 1.

So, p(1) = (1)4 + (1)3 – 2(1)2 + 1 + 1

= 2

So, by the Remainder Theorem, 2 is the remainder when x4 + x3 – 2x2 + x + 1 is divided by x – 1.


Example 10 : Check whether the polynomial q(t) = 4t3 + 4t2– t – 1 is a multiple of 2t + 1.

Solution : As you know, q(t) will be a multiple of 2t + 1 only, if 2t + 1 divides q(t) leaving remainder zero. Now, taking 2t + 1 = 0, we have t = –1 /2.

Also,

q(-1 /2) = 4(-1 /2) 3 + 4(-1 /2) 2– (-1 /2)  – 1

= –1 /2 + 1 + 1 /2 – 1

= 0

So the remainder obtained on dividing q(t) by 2t + 1 is 0.

So, 2t + 1 is a factor of the given polynomial q(t), that is q(t) is a multiple of 2t + 1.


Polynomial – Exercise 2.3 – Class IX


Factorisation of Polynomials

Let us now look at the situation of Example 10 above more closely. It tells us that since the remainder, q(-1/2) = 0, (2t + 1) is a factor of q(t), i.e., q(t) = (2t + 1) g(t) for some polynomial g(t). This is a particular case of the following theorem.

Factor Theorem : If p(x) is a polynomial of degree n > 1 and a is any real number, then

(i) x – a is a factor of p(x), if p(a) = 0, and

(ii) p(a) = 0, if x – a is a factor of p(x).

This actually follows immediately from the Remainder Theorem, but we shall not prove it here. However, we shall apply it quite a bit, as in the following examples.

 

Example 11 : Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6 and of 2x + 4.

Solution : The zero of x + 2 is –2. Let p(x) = x3 + 3x2 + 5x + 6 and s(x) = 2x + 4

Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6

= –8 + 12 – 10 + 6

= 0

So, by the Factor Theorem, x + 2 is a factor of x3 + 3x2 + 5x + 6.

Again, s(–2) = 2(–2) + 4 = 0

So, x + 2 is a factor of 2x + 4. In fact, you can check this without applying the Factor Theorem, since 2x + 4 = 2(x + 2).

 

Example 12 : Find the value of k, if x – 1 is a factor of 4x3 + 3x2 – 4x + k.

Solution : As x – 1 is a factor of p(x) = 4x3 + 3x2 – 4x + k, p(1) = 0

Now, p(1) = 4(1)3+ 3(1)2 – 4(1) + k

So, 4 + 3 – 4 + k = 0 i.e., k = –3

We will now use the Factor Theorem to factorise some polynomials of degree 2 and 3.

You are already familiar with the factorisation of a quadratic polynomial like x2+ lx + m. You had factorised it by splitting the middle term lx as ax + bx so that ab = m. Then x2+ lx + m = (x + a) (x + b). We shall now try to factorise quadratic polynomials of the type ax2+ bx + c, where a ≠ 0 and a, b, c are constants.

Factorisation of the polynomial ax2 + bx + c by splitting the middle term is as follows:

Let its factors be (px + 1) and (rx + s). Then,

ax2 + bx + c = (px + q) (rx + s) = pr x2 + (ps + qr) x + qs

Comparing the coefficients of x2 , we get a = pr.

Similarly, comparing the coefficients of x, we get b = ps + qr.

And, on comparing the constant terms, we get c = qs.

This shows us that b is the sum of two numbers ps and qr, whose product is (ps)(qr) = (pr)(qs) = ac.

Therefore, to factorise ax2 + bx + c, we have to write b as the sum of two numbers whose product is ac. This will be clear from Example 13.

 

Example 13 : Factorise 6x2 + 17x + 5 by splitting the middle term, and by using the Factor Theorem.

Solution 1 : (By splitting method) : If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.

So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us p + q = 17.

So, 6x2+ 17x + 5 = 6x2 + (2 + 15)x + 5

= 6x2+ 2x + 15x + 5

= 2x(3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Solution 2 : (Using the Factor Theorem)

6x2 + 17x + 5 = 6(x2 + 17/6x + 5/6) = 6p(x), say.

If a and b are the zeroes of p(x), then 6x2 + 17x + 5 = 6(x – a)(x – b). So, ab = 5/6 . Let s look at some possibilities for a and b. They could be ±1/2, ±1/3, ±5/3, ±5/2, ±1. Now p(1/2) = 1/4 + 17/6*(1/) + 5/6 ≠0. But p(-1/3) = 0. So, (x + 1/3) is a factor of p(x). Similarly, by trial, you can find that (x + 5/2) is a factor of p(x)

Therefore,

6x2 + 17x + 5 = 6(x + 1/3)(x + 5/2)

= 6(3x+1/3)(2x+5/2)

= (3x + 1)(2x + 5)

For the example above, the use of the splitting method appears more efficient. However, let us consider another example.

 

Example 14 : Factorise y2 – 5y + 6 by using the Factor Theorem.

Solution : Let p(y) = y2– 5y + 6. Now, if p(y) = (y – a) (y – b), you know that the constant term will be ab. So, ab = 6. So, to look for the factors of p(y), we look at the factors of 6.

The factors of 6 are 1, 2 and 3.

Now, p(2) = 22 – (5 × 2) + 6 = 0

So, y – 2 is a factor of p(y).

Also, p(3) = 32– (5 × 3) + 6 = 0

So, y – 3 is also a factor of y2– 5y + 6.

Therefore, y2– 5y + 6 = (y – 2)(y – 3)

Note that y2– 5y + 6 can also be factorised by splitting the middle term –5y.

Now, let us consider factorising cubic polynomials. Here, the splitting method will not be appropriate to start with. We need to find at least one factor first, as you will see in the following example.

 

Example 15 : Factorise x3– 23x2+ 142x – 120.

Solution : Let p(x) = x3– 23x2 + 142x – 120

We shall now look for all the factors of –120. Some of these are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.

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

Now we see that x3– 23x2+ 142x – 120 = x3– x2 – 22x2+ 22x + 120x – 120

= x2(x –1) – 22x(x – 1) + 120(x – 1) (Why?)

= (x – 1) (x2– 22x + 120) [Taking (x – 1) common]

We could have also got this by dividing p(x) by x – 1.

Now x2– 22x + 120 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:

x2– 22x + 120 = x2– 12x – 10x + 120

= x(x – 12) – 10(x – 12)= (x – 12) (x – 10)

So, x3– 23x2 – 142x – 120 = (x – 1)(x – 10)(x – 12)


Polynomials – Exercise 2.4


Algebraic Identities:

From your earlier classes, you may recall that an algebraic identity is an algebraic equation that is true for all values of the variables occurring in it. You have studied the following algebraic identities in earlier classes:

Identity I : (x + y)2 = x2 + 2xy + y2

Identity II : (x – y)2= x2 – 2xy + y2

Identity III : x2– y2 = (x + y) (x – y)

Identity IV : (x + a) (x + b) = x2 + (a + b)x + ab

You must have also used some of these algebraic identities to factorise the algebraic expressions. You can also see their utility in computations.

Example 16 : Find the following products using appropriate identities:

(i) (x + 3) (x + 3)

(ii) (x – 3) (x + 5)

Solution : (i) Here we can use Identity I : (x + y)2= x2+ 2xy + y2.

Putting y = 3 in it, we get (x + 3) (x + 3) = (x + 3)2 = x2+ 2(x)(3) + (3)2= x2+ 6x + 9

(ii) Using Identity IV above, i.e., (x + a) (x + b) = x2+ (a + b)x + ab, we have

(x – 3) (x + 5) = x2+ (–3 + 5)x + (–3)(5) = x2 + 2x – 15

 

Example 17 : Evaluate 105 × 106 without multiplying directly.

Solution : 105 × 106 = (100 + 5) × (100 + 6)

= (100)2 + (5 + 6) (100) + (5 × 6), using Identity IV

= 10000 + 1100 + 30 = 11130

You have seen some uses of the identities listed above in finding the product of some given expressions. These identities are useful in factorisation of algebraic expressions also, as you can see in the following examples.

Example 18 : Factorise:

(i) 49a2+ 70ab + 25b2

(ii)25/4 x21/9 y2

Solution : (i) Here you can see that

49a2= (7a)2, 25b2= (5b)2, 70ab = 2(7a) (5b)

Comparing the given expression with x2+ 2xy + y2, we observe that x = 7a and y = 5b.

Using Identity I, we get

49a2+ 70ab + 25b2= (7a + 5b)2= (7a + 5b) (7a + 5b)

(ii) We have 25/4 x21/9 y2 = (5/2x)2 – (y/3)2

Now comparing it with identity III, we get

25/4 x21/9 y2 = (5/2x)2 – (y/3)2

= (5/2 x + y/3) (5/2 x – y/3)

So far, all our identities involved products of binomials. Let us now extend the Identity I to a trinomial x + y + z. We shall compute (x + y + z)2 by using Identity I.

Let x + y = t. Then,(x + y + z)2= (t + z)2

= t2+ 2tz + t2 (Using Identity I)

= (x + y)2+ 2(x + y)z + z2 (Substituting the value of t)

= x2+ 2xy + y2+ 2xz + 2yz + z2 (Using Identity I)

= x2+ y2+ z2+ 2xy + 2yz + 2zx (Rearranging the terms)

So, we get the following identity:

Identity V : (x + y + z)2= x2+ y2+ z2+ 2xy + 2yz + 2zx

 

Remark : We call the right hand side expression the expanded form of the left hand side expression. Note that the expansion of (x + y + z)2 consists of three square terms and three product terms.

 

Example 19 : Write (3a + 4b + 5c)2 in expanded form.

Solution : Comparing the given expression with (x + y + z)2

, we find that  x = 3a, y = 4b and z = 5c.

Therefore, using Identity V, we have

(3a + 4b + 5c)2

= (3a)2+ (4b)2+ (5c)2+ 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)

= 9a2+ 16b2+ 25c2+ 24ab + 40bc + 30ac

 

Example 20 : Expand (4a – 2b – 3c)2.

Solution : Using Identity V, we have (4a – 2b – 3c)2

= [4a + (–2b) + (–3c)]2

= (4a)2 + (–2b)2+ (–3c)2+ 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a)

= 16a2+ 4b2+ 9c2– 16ab + 12bc – 24ac

 

Example 21 : Factorise 4x2+ y2+ z2– 4xy – 2yz + 4xz.

Solution : We have 4x2+ y2+ z2– 4xy – 2yz + 4xz = (2x)2+ (–y)2+ (z)2+ 2(2x)(–y)+ 2(–y)(z) + 2(2x)(z) = [2x + (–y) + z]2 (Using Identity V)

= (2x – y + z)2= (2x – y + z)(2x – y + z)

So far, we have dealt with identities involving second degree terms. Now let us extend Identity I to compute (x + y)3

We have, (x + y)3= (x + y) (x + y)2

= (x + y)(x2+ 2xy + y2)

= x(x2+ 2xy + y2) + y(x2+ 2xy + y2)

= x3+ 2x2y + xy2+ x2y + 2xy2+ y3

= x3+ 3x2y + 3xy2+ y3

= x3+ y3+ 3xy(x + y)

So, we get the following identity:

Identity VI : (x + y)3= x3+ y3+ 3xy (x + y)

Also, by replacing y by –y in the Identity VI, we get Identity VII : (x – y)3= x3– y3– 3xy(x – y)

= x3 – 3x2y + 3xy2 – y3

 

Example 22 : Write the following cubes in the expanded form:

(i) (3a + 4b)3

(ii) (5p – 3q)3

Solution : (i) Comparing the given expression with (x + y)3, we find that x = 3a and y = 4b.

So, using Identity VI, we have:

(3a + 4b)3= (3a)3+ (4b)3+ 3(3a)(4b)(3a + 4b)

= 27a3+ 64b3+ 108a2b + 144ab2

 

(ii) Comparing the given expression with (x – y)3, we find that x = 5p, y = 3q.

So, using Identity VII, we have:

(5p – 3q)3= (5p)3– (3q)3 – 3(5p)(3q)(5p – 3q)

= 125p3– 27q3– 225p2q + 135pq2

 

Example 23 : Evaluate each of the following using suitable identities:

(i) (104)3 (ii) (999)3

Solution : (i) We have, (104)3

= (100 + 4)3

= (100)3+ (4)3+ 3(100)(4)(100 + 4) (Using Identity VI)

= 1000000 + 64 + 124800

= 1124864

(ii) We have, (999)3= (1000 – 1)3

= (1000)3– (1)3– 3(1000)(1)(1000 – 1)(Using Identity VII)

= 1000000000 – 1 – 2997000

= 997002999

 

Example 24 : Factorise 8x3+ 27y3+ 36x2y + 54xy2

Solution : The given expression can be written as

(2x)3+ (3y)3 + 3(4x2)(3y) + 3(2x)(9y2)

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

(3y) + 3(2x)(3y)2

= (2x + 3y)3 (Using Identity VI)

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

Now consider (x + y + z)(x2+ y2 + z2– xy – yz – zx)

On expanding, we get the product as

x(x2+ y2+ z2– xy – yz – zx) + y(x2+ y2+ z2– xy – yz – zx)+ z(x2+ y2+ z2– xy – yz – zx) = x3+ xy2+ xz2– x2y – xyz – zx2+ x2y+ y3+ yz2– xy2– y2z – xyz + x2z + y2z + z3– xyz – yz2– xz2

= x3+ y3+ z3– 3xyz (On simplification)

 

So, we obtain the following identity:

Identity VIII : x3+ y3+ z3– 3xyz = (x + y + z)(x2+ y2+ z2– xy – yz – zx)

 

Example 25 : Factorise : 8x3+ y3+ 27z3– 18xyz

Solution : Here, we have, 8x3+ y3+ 27z3 – 18xyz

= (2x)3+ y3+ (3z)3– 3(2x)(y)(3z)

= (2x + y + 3z)[(2x)2+ y2+ (3z)2– (2x)(y) – (y)(3z) – (2x)(3z)]

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


Polynomial – Exercise 2.5


Advertisements

One thought on “Polynomials – Class IX [CBSE]

  1. Pingback: IX – Table of Contents | Breath Math

Comments are closed.