Polynomials And Factorisation

Degree of p(x) is the highest power of x in p(x).

(i) The highest power of x in x⁵ - x⁴ + 3 is 5.

∴ The degree of x⁵ - x⁴ + 3 is 5.

(ii) The highest power of x in x² + x - 5 is 2.

∴ The degree of x² + x - 5 is 2.

(iii) The highest power of x in 5 is 0(∵ there is no term of x).

∴ The degree of 5 is 0.

(iv) The highest power of x in 3x⁶ + 6y³ - 7 is 6.

∴ The degree of 3x⁶ + 6y³ - 7 is 6.

(v) The highest power of y in 4 - y² is 2.

∴ The degree of 4 - y² is 2.

(vi) The highest power of t in 5t – √3 is 1.

∴ The degree of 5t – √3 is 1.

(i) 3x² - 2x + 5 has only one variable that is x.

∴ yes, it is a polynomial in one variable.

(ii) x² + √2 has only one variable that is x.

∴ yes, it is a polynomial in one variable.

(iii) p² – 3p + q has two variables that are p and q.

∴ no, it is not a polynomial in one variable.

(iv) has a negative exponent of y.

∴ no, it is not a polynomial.

(v) The exponent of x in 5√x + x√5 is 1/2 which is not a non-negative integer

∴ no, it is not a polynomial.

(vi) x¹⁰⁰ + y¹⁰⁰ has two variables that are x and y.

∴ no, it is not a polynomial in one variable.

A coefficient is a multiplicative factor in some term of a polynomial. It is the constant written before the variable.

Therefore,

(i) The constant written before x³ in x³ + x + 1 is 1.

∴ The coefficient of x³ in x³ + x + 1 is 1.

(ii) The constant written before x³ in 2 – x³ + x² is -1.

∴ The coefficient of x³ in 2 – x³ + x² is -1.

(iii) The constant written before x³ in √2x³ + 5 is √2.

∴ The coefficient of x³ in √2x³ + 5 is √2.

(iv) The constant written before x³ in 2x³ + 5 is 2.

∴ The coefficient of x³ in 2x³ + 5 is 2.

(v) The constant written before x³ in is.

∴ The coefficient of x³ in is.

(vi) The constant written before x³ in is.

∴ The coefficient of x³ in is.

(vii) The term x³ does not exist in 2x² + 5.

∴ The coefficient of x³ in 2x² + 5 is 0.

(viii) The term x³ does not exist in 4.

∴ The coefficient of x³ in 4 is 0.

(i) A quadratic polynomial is a polynomial of degree 2

∵ the degree of 5x² + x – 7 is 2

∴ 5x² + x – 7 is a quadratic polynomial.

(ii) A cubic polynomial is a polynomial of degree 3

∵ the degree of 5x² + x – 7 is 3

∴ 5x² + x – 7 is a cubic polynomial.

(iii) A quadratic polynomial is a polynomial of degree 2

∵ the degree of x² + x + 4 is 2

∴ x² + x + 4 is a quadratic polynomial.

(iv) A linear polynomial is a polynomial of degree 1

∵ the degree of x – 1 is 1

∴ x – 1 is a linear polynomial.

(v) A linear polynomial is a polynomial of degree 1

∵ the degree of 3p is 1

∴ 3p is a linear polynomial.

(vi) A quadratic polynomial is a polynomial of degree 2

∵ the degree of πr² is 2

∴ πr² is a quadratic polynomial.

(i) A polynomial with two terms is called a binomial.

∴ The statement is true.

(ii) A polynomial can have more than two terms.

∴ The statement is false.

(iii) A binomial should have two terms, the degree of those terms can be any integer.

∴ The statement is true.

(iv) The constant polynomial whose coefficients are all equal to 0, is called a zero polynomial. Its degree can be any integer.

∴ The statement is false.

(v) The highest power in x² + 2xy + y² is 2, therefore its degree is 2.

∴ The statement is true.

(vi) A monomial is a polynomial which has only one term.

∵ πr² has only one term

∴ The statement is true.

A monomial is a polynomial which has only one term, and the degree is the highest power of the variable. Therefore, an example of a monomial of degree 10 is 3x¹⁰.

A trinomial is a polynomial which has three terms, and the degree is the highest power of the variable. Therefore, example of a trinomial of degree 10 is 3x¹⁰ + 2x² + 5.

(i) p(x) = 4x² – 5x + 3

⇒ p(0) = 4(0)² – 5(0) + 3

⇒ p(0) = 0– 0 + 3

⇒ p(0) = 3

(ii) p(x) = 4x² – 5x + 3

⇒ p(-1) = 4(-1)² – 5(-1) + 3

⇒ p(-1) = 4 × 1– (-5) + 3

⇒ p(-1) = 4 +5 + 3

⇒ p(-1) = 12

(iii) p(x) = 4x² – 5x + 3

⇒ p(2) = 4(2)² – 5(2) + 3

⇒ p(2) = 4 × 4– 10 + 3

⇒ p(2) = 16– 10 + 3

⇒ p(2) = 9

(iv) p(x) = 4x² – 5x + 3

p(x) = x² – x + 1

⇒ p(0) = (0)² – 0 + 1

⇒ p(0) = 1

And,

⇒ p(1) = (1)² – 1 + 1

⇒ p(1) = 1– 1 + 1

⇒ p(1) = 1

And,

⇒ p(2) = (2)² – 2 + 1

⇒ p(2) = 4– 2 + 1

⇒ p(2) = 3

p(y) = 2 + y + 2y² – y³

⇒ p(0) = 2 + 0 + 2(0)² – (0)³

⇒ p(0) = 2 + 0 + 0– 0

⇒ p(0) = 2

And,

⇒ p(1) = 2 + 1 + 2(1)² – (1)³

⇒ p(1) = 2 + 1 + 2– 1

⇒ p(1) = 4

And,

⇒ p(2) = 2 + 2 + 2(2)² – (2)³

⇒ p(2) = 2 + 2 + 8– 8

⇒ p(2) = 4

p(z) = z³

⇒ p(0) = 0³

⇒ p(0) = 0

And,

⇒ p(1) = 1³

⇒ p(1) = 1

And,

⇒ p(2) = 2³

⇒ p(2) = 8

p(t) = (t - 1) (t + 1)

⇒ p(t) = t² – 1

⇒ p(0) = (0)² – 1

⇒ p(0) = 0– 1

⇒ p(0) = -1

And,

⇒ p(1) = (1)² – 1

⇒ p(1) = 1– 1

⇒ p(1) = 0

And,

⇒ p(2) = (2)² – 1

⇒ p(2) = 4– 1

⇒ p(2) = 3

p(t) = x² - 3x + 2

⇒ p(0) = (0)² – 3(0) + 2

⇒ p(0) = 0– 0 + 2

⇒ p(0) = 2

And,

⇒ p(1) = (1)² – 3(1) + 2

⇒ p(1) = 1– 3 + 2

⇒ p(0) = 0

And,

⇒ p(2) = (2)² – 3(2) + 2

⇒ p(2) = 4– 6 + 2

⇒ p(2) = 0

p(x) = 2x + 1

⇒ p(-1/2) = 2(-1/2) + 1

⇒ p(-1/2) = -1 + 1

⇒ p(-1/2) = 0

∴ Yes x = -1/2 is the zero of polynomial 2x + 1.

p(x) = 5x - π

∴ No x = - is not the zero of polynomial 5x – π.

p(x) = x² - 1

⇒ p(-1) = (-1)² – 1

⇒ p(-1) = 1 – 1

⇒ p(-1) = 0

∴ Yes x = -1 is the zero of polynomial x² – 1.

And,

⇒ p(1) = (1)² – 1

⇒ p(1) = 1 – 1

⇒ p(1) = 0

∴ Yes x = 1 is the zero of polynomial x² – 1.

p(x) = (x - 1)(x + 2)

⇒ p(-1) = (-1 - 1)(-1 + 2);

⇒ p(-1) = -2 × 1

⇒ p(-1) = -2

∴ No x = -1 is not the zero of polynomial (x - 1)(x + 2).

And,

⇒ p(-2) = (-2 - 1)(-2 + 2);

⇒ p(-2) = -3 × 0

⇒ p(-2) = 0

∴ Yes x = -2 is not the zero of polynomial (x - 1)(x + 2).

p(y) = y²

⇒ p(0) = 0²

⇒ p(0) = 0

∴ Yes y = 0 is the zero of polynomial y².

p(x) = ax + b

∴ Yes is the zero of polynomial ax + b.

f(x) = 3x² - 1

∴ Yes is the zero of polynomial 3x² - 1.

∴ No is not the zero of polynomial 3x² - 1.

f(x) = 2x - 1

∴ Yes is the zero of polynomial 2x - 1.

∴ No is not the zero of polynomial 2x - 1.

f(x) = x + 2

f(x) = 0

⇒ x + 2 = 0

⇒ x = 0 – 2

⇒ x = -2

∴ x = -2 is the zero of the polynomial x + 2.

f(x) = x - 2

f(x) = 0

⇒ x – 2 = 0

⇒ x = 0 + 2

⇒ x = 2

∴ x = 2 is the zero of the polynomial x – 2.

f(x) = 2x + 3

f(x) = 0

⇒ 2x + 3 = 0

⇒ 2x = 0 – 3

⇒ 2x = -3

∴ is the zero of the polynomial 2x + 3.

f(x) = 2x – 3

f(x) = 0

⇒ 2x – 3 = 0

⇒ 2x = 0 + 3

⇒ 2x = 3

∴ is the zero of the polynomial 2x – 3.

f(x) = x²

f(x) = 0

⇒ x² = 0

⇒ x = 0

∴ x = 0 is the zero of the polynomial x².

f(x) = px, p ≠ 0

f(x) = 0

⇒ px = 0

⇒ x = 0

∴ x = 0 is the zero of the polynomial px.

f(x) = px + q, p ≠ 0, p q are real numbers.

f(x) = 0

⇒ px + q = 0

⇒ px = -q

∴ is the zero of the polynomial px + q.

∵ 2 is the zeroes of the polynomial p(x) = 2x² - 3x + 7a

∴ p(2) = 0

Now,

p(x) = 2x² - 3x + 7a

⇒p(2) = 2(2)² – 3(2)+ 7a

⇒ 2 × 4– 3 × 2 + 7a = 0

⇒ 8– 6 + 7a = 0

⇒ 2 + 7a = 0

⇒ 7a = -2

∵ 0 and 1 are the zeroes of the polynomial f(x) = 2x³-3x²+ ax + b

∴ f(0) = 0 and f(1) = 1

Now,

f(x) = 2x³-3x²+ ax + b

⇒ f(0) = 2(0)³ – 3(0)²+ a(0) + 0

⇒ 2 × 0– 3 × 0 + a × 0 + b = 0

⇒ 0– 0 + 0 + b = 0

⇒ b = 0

And,

⇒ f(1) = 2(1)³ – 3(1)²+ a(1) +1

⇒ 2 × 1– 3 × 1 + a × 1 + b = 0

⇒ 2– 3 + a + b = 0

⇒ 2– 3 + a + 0 = 0 [∵ b = 0]

⇒ -1 + a = 0

⇒ a = 1

Let p(x) = x³ + 3x² + 3x + 1

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x+1 is p(–1)

p(–1) = (–1)³ + 3(–1)² + 3(–1) +1

⇒ p(–1) = –1 + 3 – 3 + 1 = 0

∴ Remainder of x³ + 3x² + 3x + 1 when divided by x+1 is 0

Let p(x) = x³ + 3x² + 3x + 1

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by is p(1/2)

∴ Remainder of x³ + 3x² + 3x + 1 when divided by is

Let p(x) = x³ + 3x² + 3x + 1

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x is p(0)

P(0) = (0)³+ 3(0)²+ 3(0) + 1

= 1

∴ Remainder of x³ + 3x² + 3x + 1 when divided by x is 1

Let p(x) = x³ + 3x² + 3x + 1

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x + π is p(–π)

p(–π) = (–π)³+ 3(–π)²+ 3(–π) + 1

⇒ p(–π)= –π³ + 3π² –3π + 1

∴ Remainder of x³ + 3x² + 3x + 1 when divided by x+ π is –π³ + 3π² –3π + 1

Let p(x) = x³ + 3x² + 3x + 1

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by 5 + 2x is

∴ Remainder of x³ + 3x² + 3x + 1 when divided by 5 + 2x is

Let q(x) = x³ – px² + 6x – p

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of q(x) when divided by x – p is q(p)

q(p) = (p)³– p(p)²+ 6(p) – p

⇒ q(p) = p³ – p³ + 6p – p

∴ Remainder of x³ – px² + 6x – p when divided by x – p is 5p

Let p(x) = 2x² – 3x + 5

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by 2x – 3 is

= 5

⇒ Remainder of 2x² – 3x + 5 when divided by 2x – 3 is 5.

As on dividing the given polynomial by 2x – 3, we get a non–zero remainder, therefore, 2x – 3 does not completely divide the polynomial.

∴ It is not a factor.

Let p(x) = 9x³ – 3x² + x – 5

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by is

⇒ Remainder of 9x³ – 3x² + x – 5 when divided by is –3

Let p(x) = 2x³ + ax² + 3x – 5 and q(x) = x³ + x² – 4x + a

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x – 2 is p(2). Similarly, Remainder of q(x) when divided by x – 2 is q(2)

⇒ p(2) = 2(2)³ +a(2)² + 3(2) – 5

⇒p(2) = 16 + 4a +6 – 5

⇒p(2) = 17 + 4a

Similarly, q(2) = (2)³ + (2)² + –4(2) + a

⇒ q(2) = 8 + 4 –8 + a

⇒ q(2) = 4 + a

Since they both leave the same remainder, so p(2) = q(2)

⇒ 17 + 4a = 4 + a

⇒ 13 = 3a

∴ The value of a is –13/3

Let p(x) = x³ + ax² + 5 and q(x) = x³ – 2x² + a

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x + 2 is p(–2). Similarly, Remainder of q(x) when divided by x + 2 is q(–2)

⇒ p(–2) = (–2)³ +a(–2)² + 5

⇒p(–2) = –8 + 4a + 5

⇒p(–2) = –3 + 4a

Similarly, q(–2) = (–2)³ – 2(–2)² + a

⇒ q(–2) = –8 –8 + a

⇒ q(–2) = –16 + a

Since they both leave the same remainder, so p(–2) = q(–2)

⇒ –3 + 4a = –16 + a

⇒ –13 = 3a

∴ The value of a is –13/3

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

Therefore, remainder when f(x) is divided by g(x) is f(2)

f(2) = 2⁴ – 3(2)² + 4

⇒ f(2) = 16 – 12 + 4 = 8

∴ the remainder when x⁴ – 3x² + 4 is divided by x – 2 is 8

Given: p(x) = x³ – 6x² + 14x – 3 and g(x) = 1 – 2x

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

Therefore, remainder when p(x) is divided by g(x) is

∴ the remainder when x³ – 6x² + 14x – 3 is divided by 1 – 2x is
Result Verification:

We can see that the remainder is .

Hence, verified.

Let p(x) = 2x³ +3x² + ax + b

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x – 2 is p(2)

p(2) = 2(2)³ +3(2)² + a(2) + b

⇒ p(2) = 16 + 12 + 2a + b

Also, it is given that p(2) = 2, on substituting value above, wev get,

2 = 28 + 2a + b

⇒ 2a + b = –26 ---------- (A)

Similarly,

Remainder of p(x) when divided by x + 2 is p(–2)

p(–2) = 2(–2)³ +3(–2)² + a(–2) + b

⇒ p(–2) = –16 + 12 – 2a + b

Also, it is given that p(–2) = –2, on substituting value above, we get,

–2 = –4 – 2a + b

⇒ – 2a + b = 2 ---------- (B)

On solving the above two equ. (A) and (b), we get,

a = –7 and b = –12

∴ Value of a and b is –7 and –12 respectively.

Let f(x) = x³ – x² – x + 1

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

⇒ f(–1) = (–1)³ – (–1)² – (–1) + 1

⇒ f(–1) = –1 –1 +1 +1

⇒ f(–1) = 0

As, f(–1) is equal to zero, therefore (x+1) is a factor x³ – x² – x + 1

Let f(x) = x⁴ – x³ + x² – x + 1

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

⇒ f(–1) = (–1)⁴ – (–1)³ + (–1)² – (–1) + 1

⇒ f(–1) = 1 + 1 + 1 + 1 + 1

⇒ f(–1) = 5

As, f(–1) is not equal to zero, therefore (x+1) is not a factor x⁴ – x³ + x² – x + 1

Let f(x) = x⁴ + 2x³ + 2x² + x + 1

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

⇒ f(–1) = (–1)⁴ + 2(–1)³ + 2(–1)² + (–1) + 1

⇒ f(–1) = 1 – 2 + 2 – 1 + 1

⇒ f(–1) = 1

As, f(–1) is not equal to zero, therefore (x+1) is not a factor x⁴ + 2x³ + 2x² + x + 1

Let f(x) = x³ – x² – (3 –√3 ) x + √3

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

⇒ f(–1) = (–1)³ – (–1)² – (3 –√3)(–1) + √3

⇒ f(–1) = –1 – 1 + 3 – √3 + √3

⇒ f(–1) = 1

As, f(–1) is not equal to zero, therefore (x+1) is not a factor x³ – x² –(3 –√3) x + √3

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

⇒ f(–1) = 5 (–1)³ + (–1)² – 5(–1) – 1

⇒ f(–1) = –5 + 1 + 5 – 1

⇒ f(–1) = 0

As, f(–1) is equal to zero, therefore, g(x) = (x+1) is a factor f(x)

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0

For checking (x+1) to be a factor, we will find f(–1)

⇒ f(–1) = (–1)³ + 3(–1)² + 3(–1) + 1

⇒ f(–1) = – 1 + 3 – 3 + 1

⇒ f(–1) = 0

As, f(–1) is equal to zero, therefore, g(x) = (x+1) is a factor f(x)

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0

For checking (x – 2) to be a factor, we will find f(2)

⇒ f(2) = (2)³ – 4(2)² + (2) + 6

⇒ f(–1) = 8 – 16 + 2 + 6

⇒ f(–1) = 0

As, f(–1) is equal to zero, therefore, g(x) = (x – 2) is a factor of f(x)

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0

For checking (3x – 2) to be a factor, we will find f(2/3)

As, f(–1) is equal to zero, therefore, g(x) = (3x – 2) is a factor of f(x)

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x – a) is a factor of p(x), if p(a) = 0

For checking (2x + 3) to be a factor, we will find f(–3/2)

As, f(–3/2) is equal to zero, therefore, g(x) = (3x – 2) is a factor of f(x)

Let f(x) = x³ – 3x² – 10x + 24

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0

For checking (x – 2) to be a factor, we will find f(2)

⇒ f(2) = (2)³ – 3(2)² – 10(2) + 24

⇒ f(2) = 8 – 12 – 20 + 24

⇒ f(2) = 0

So, (x–2) is a factor.

For checking (x + 3) to be a factor, we will find f(–3)

⇒ f(–3) = (–3)³ – 3(–3)² – 10(–3) + 24

⇒ f(–3) = –27 – 27 + 30 + 24

⇒ f(–3) = 0

So, (x+3) is a factor.

For checking (x – 4) to be a factor, we will find f(4)

⇒ f(4) = (4)³ – 3(4)² – 10(4) + 24

⇒ f(4) = 64 – 48 – 40 + 24

⇒ f(4) = 0

So, (x–4) is a factor.

∴ (x – 2), (x + 3) and (x – 4) are factors of x³ – 3x² – 10x + 24

Let f(x) = x³ – 6x² – 19x + 84

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0

For checking (x + 4) to be a factor, we will find f(–4)

⇒ f(–4) = (–4)³ – 6(–4)² – 19(–4) + 84

⇒ f(–4) = –64 – 96 + 76 + 84

⇒ f(–4) = 0

So, (x+4) is a factor.

For checking (x – 3) to be a factor, we will find f(3)

⇒ f(3) = (3)³ – 6(3)² – 19(3) + 84

⇒ f(3) = 27 – 54 – 57 + 84

⇒ f(3) = 0

So, (x–3) is a factor.

For checking (x – 7) to be a factor, we will find f(7)

⇒ f(7) = (7)³ – 6(7)² – 19(7) + 84

⇒ f(7) = 343 – 294 – 133 + 84

⇒ f(7) = 0

So, (x–7) is a factor.

∴ (x + 4), (x – 3) and (x – 7) are factors of x³ – 3x² – 10x + 24

Let f(x) = px² + 5x + r

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.

So, if (x – 2) is a factor of f(x)

⇒ f(2) = 0

⇒ p(2)² + 5(2) + r = 0

⇒ 4p + r = –10 -------- (A)

Also as is also a factor,

⇒ p + 4r = –10 ----- (B)

Let f(x) = ax⁴ + bx³ + cx² + dx + e

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.

Also we can write, (x² – 1) = (x + 1)(x – 1)

Since (x² – 1) is a factor of f(x), this means (x + 1) and (x – 1) both are factors of f(x).

So, if (x – 1) is a factor of f(x)

⇒ f(1) = 0

⇒ a(1)⁴ + b(1)³ + c(1)² + d(1) + e = 0

⇒ a + b + c + d + e = 0 ----- (A)

Also as (x + 1) is also a factor,

⇒ f(–1) = 0

⇒ a(–1)⁴ + b(–1)³ + c(–1)² + d(–1) + e = 0

⇒ a – b + c – d + e = 0

⇒ a + c + e = b + d ---- (B)

On solving equations (A) and (B), we get,

a + c + e = b + d = 0

Let p(x) = x³ – 2x² – x + 2

By trial, we find that p(1) = 0, so by Factor theorem,

(x – 1) is the factor of p(x)

When we divide p(x) by (x – 1), we get x² – x – 2.

Now, (x² – x – 2) is a quadratic and can be solved by splitting the middle terms.

We have x² – x – 2 = x² – 2x + x – 2

⇒ x (x – 2) + 1 (x – 2)

⇒ (x + 1)(x – 2)

So, x³ – 2x² – x + 2 = (x – 1)(x + 1)(x – 2)

Let p(x) = x³ – 3x² – 9x – 5

By trial, we find that p(–1) = 0, so by Factor theorem,

(x + 1) is the factor of p(x)

When we divide p(x) by (x + 1), we get x² – 4x – 5.

Now, (x² – 4x – 5) is a quadratic and can be solved by splitting the middle terms.

We have x² – 4x – 5 = x² – 5x + x – 5

⇒ x (x – 5) + 1 (x – 5)

⇒ (x + 1)(x – 5)

So, x³ – 3x² – 9x – 5= (x + 1)(x + 1)(x – 5)

Let p(x) = x³ + 13x² + 32x + 20

By trial, we find that p(–1) = 0, so by Factor theorem,

(x + 1) is the factor of p(x)

When we divide p(x) by (x + 1), we get x² + 12x + 20.

Now, (x² + 12x + 20) is a quadratic and can be solved by splitting the middle terms.

We have x² + 12x + 20= x² + 10x + 2x + 20

⇒ x (x + 10) + 2 (x + 10)

⇒ (x + 2)(x + 10)

So, x³ + 13x² + 32x + 20 = (x + 1)(x + 2)(x + 10)

Let p(y) = y³ + y² – y – 1

On taking y² common from first two terms in p(y), we get,

p (y) = y²(y + 1) –1(y + 1)

Now, taking (y + 1) common, we get,

⇒ p(y) = (y² – 1)(y + 1)

As we know the identity, (y² – 1) = (y + 1)(y – 1)

⇒ p(y) = (y – 1)(y + 1)(y + 1)

Let f(x) = ax² + bx + c and p(x) = bx² + ax + c

As (x + 1) is the common factor of f(x) and p(x) both, and as by Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.

⇒ f(–1) = p (–1) = 0

⇒ a(–1)² + b(–1) + c = b(–1)² + a(–1) + c

⇒ a – b + c = b – a + c

⇒ 2a = 2b

⇒ a = b ------ (A)

Also, we discussed that,

f(–1) = 0

⇒ a(–1)² + b(–1) + c = 0

⇒ a – b + c = 0

From equation (A), we see that a = b,

⇒ c = 0 -------- (B)

∴ Equations (A) and (B) show us the required result.

Let f(x) = x² – x – 6 and p(x) = x² + 3x – 18

As (x – a) is the common factor of f(x) and p(x) both, and as by Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.

⇒ f(a) = p (a)

⇒ (a)² – (a) – 6 = (a)² + 3(a) – 18

⇒ 4a = 12

⇒ a = 3

∴ The value of a is 3.

Let f(x) = y³ – 2y² – 9y + 18

Taking y² common from the first two terms of f(x) and 9 from the last two terms of f(x), we get,

⇒ f(x) = y²(y – 2) –9(y – 2)

Now, taking (y – 2) common from above,

⇒ f(x) = (y² – 9)(y – 2) -------- (A)

We know the identity as,

a² – b² = (a – b)(a + b)

So, using above identity on equation (A), we get,

⇒ f(x) = (y + 3)(y – 3)(y – 2)

∴ the other two factors of y³ – 2y² – 9y + 18 besides (y – 3) are (y + 3) and (y – 2).

using the identity (x + a) × (x + b) = x² + (a + b)x + ab

here a = 5 and b = 2

⇒ (x + 5) (x + 2) = x² + (5 + 2)x + 5 × 2

Therefore (x + 5) (x + 2) = x² + 7x + 10

(x - 5) (x - 5) = (x – 5)²

Using identity (a – b)² = a² – 2ab + b²

Here a = x and b = 5

⇒ (x - 5) (x - 5) = x² – 2 × x × 5 + 5²

⇒ (x - 5) (x - 5) = x² – 10x + 25

Therefore (x - 5) (x - 5) = x² – 10x + 25

using the identity (a + b) × (a – b) = a² – b²

here a = 3x and b = 2

⇒ (3x + 2)(3x - 2) = (3x)² – 2²

Therefore (3x + 2)(3x - 2) = 9x² – 4

using the identity (a + b) × (a – b) = a² – b²

here a = x² and b =

⇒ = (x²)² -

Therefore = x⁴ -

(1 + x) (1 + x) = (1 + x)²

Using identity (a + b)² = a² + 2ab + b²

Here a = 1 and b = x

⇒ (1 + x) (1 + x) = 1² + 2(1)(x) + x²

Therefore (1 + x) (1 + x) = 1 + 2x + x²

101 can be written as (100 + 1) and

99 can be written as (100 - 1)

⇒ 101 × 99 = (100 + 1) × (100 - 1)

using the identity (a + b) × (a – b) = a² – b²

here a = 100 and b = 1

⇒ 101 × 99 = 100² – 1²

⇒ 101 × 99 = 10000 – 1

⇒ 101 × 99 = 9999

999 can be written as (1000 – 1)

⇒ 999 × 999 = (1000 – 1) × (1000 – 1)

⇒ 999 × 999 = (1000 – 1)²

Using identity (a – b)² = a² – 2ab + b²

Here a = 1000 and b = 1

⇒ 999 × 999 = 1000² – 2(1000)(1) + 1²

⇒ 999 × 999 = 1000000 – 2000 + 1

⇒ 999 × 999 = 998000 + 1

⇒ 999 × 999 = 998001

⇒ 50 = and 49 =

⇒ 50 = and 49 =

⇒ 50 = and 49 =

⇒ = ×

⇒ =

Consider 101 × 99

101 can be written as (100 + 1) and

99 can be written as (100 - 1)

⇒ 101 × 99 = (100 + 1) × (100 - 1)

using the identity (a + b) × (a – b) = a² – b²

here a = 100 and b = 1

⇒ 101 × 99 = 100² – 1²

⇒ 101 × 99 = 10000 – 1

⇒ 101 × 99 = 9999

Therefore =

501 can be written as (500 + 1)

⇒ 501 × 501 = (500 + 1) × (500 + 1)

⇒ 501 × 501 = (500 + 1)²

⇒ 501 × 501 = (500 + 1) × (500 + 1)

⇒ 501 × 501 = (500 + 1)²

Using identity (a + b)² = a² + 2ab + b²

Here a = 500 and b = 1

⇒ 501 × 501 = 500² + 2(500)(1) + 1²

⇒ 501 × 501 = 250000 + 1000 + 1

⇒ 501 × 501 = 251001

30.5 = and 29.5 =

⇒ 30.5 × 29.5 = ×

⇒ 30.5 × 29.5 = ×

⇒ 30.5 × 29.5 = …(i)

Consider 61 × 59

61 = (60 + 1)

59 = (60 – 1)

⇒ 61 × 59 = (60 + 1)(60 – 1)

using the identity (a + b) × (a – b) = a² – b²

here a = 60 and b = 1

⇒ 61 × 59 = 60² – 1²

⇒ 61 × 59 = 3600 – 1

⇒ 61 × 59 = 3599

From (i)

⇒ 30.5 × 29.5 =

Therefore 30.5 × 29.5 = 899.75

16x² can be written as (4x)²

24xy can be written as 2(4x)(3y)

9y² can be written as (3y)²

⇒ 16x² + 24xy + 9y² = (4x)² + 2(4x)(3y) + (3y)²

Using identity (a + b)² = a² + 2ab + b²

Here a = 4x and b = 3y

⇒ 16x² + 24xy + 9y² = (4x + 3y)²

Therefore 16x² + 24xy + 9y² = (4x + 3y) (4x + 3y)

4y² can be written as (2y)²

4y can be written as 2(1)(2y)

1 can be written as 1²

⇒ 4y² - 4y + 1 = (2y)² - 2(1)(2y) + 1²

Using identity (a - b)² = a² - 2ab + b²

Here a = 2y and b = 1

⇒ 4y² - 4y + 1 = (2y - 1)²

Therefore 4y² - 4y + 1 = (2y - 1) (2y - 1)

4x² can be written as (2x)²

can be written as

⇒ 4x² - = (2x)² -

using the identity (a + b) × (a – b) = a² – b²

here a = 2x and b =

Therefore 4x² - = (2x + ) (2x - )

Take out common factor 2

⇒ 18a² – 50 = 2 (9a² - 25)

Now

9a² can be written as (3a)²

25 can be written as 5²

⇒ 18a² – 50 = 2 ((3a)² – 5²)

using the identity (a + b) × (a – b) = a² – b²

here a = 3a and b = 5

therefore 18a² – 50 = 2 (3a + 5) (3a – 5)

Given is quadratic equation which can be factorised by splitting the middle term as shown

⇒ x² + 5x + 6 = x² + 3x + 2x + 6

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

= (x + 3) (x + 2)

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

Take out common factor 3

⇒ 3p² - 24p + 36 = 3 (p² – 8p + 12)

Now splitting the middle term of quadratic p² – 8p + 12 to factorise it

⇒ 3p² - 24p + 36 = 3 (p² – 6p – 2p + 12)

= 3 [p (p – 6) – 2 (p – 6)]

= 3 (p – 2) (p – 6)

Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Here a = x, b = 2y and c = 4z

⇒ (x + 2y + 4z)² = x² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)

Therefore

(x + 2y + 4z)² = x² + 4y² + 16z² + 4xy + 16yz + 8xz

Using identity (x – y)³ = x³ - y³ – 3x²y + 3xy²

Here x = 2a and y = 3b

⇒ (2a - 3b)³ = (2a)³ – (3b)³ – 3(2a)²(3b) + 3(2a)(3b)²

= 8a³ – 27b³ – 18a²b + 18ab²

Therefore (2a - 3b)³ = 8a³ – 27b³ – 18a²b + 18ab²

(-2a + 5b - 3c)² = [(-2a) + (5b) + (-3c)]²

Using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Here x = -2a, y = 5b and z = -3c

⇒ (-2a + 5b - 3c)² = (-2a)² + (5b)² + (-3c)² + 2(-2a)(5b) + 2(5b)(-3c) + 2(-3c)(-2a)

⇒ (-2a + 5b - 3c)² = 4a² + 25b² + 9c² + (-20ab) + (-30bc) + 12ac

⇒ (-2a + 5b - 3c)² = 4a² + 25b² + 9c² - 20ab - 30bc + 12ac

Therefore

(-2a + 5b - 3c)² = 4a² + 25b² + 9c² - 20ab - 30bc + 12ac

= [ + + 1]²

Using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Here x = , y = and z = 1

⇒ = + + 1² + 2 + 2(1) + 2(1)

⇒ = + + 1 + + (-b) +

⇒ = + + 1 - – b +

Therefore = + + 1 - – b +

Using identity (x + y)³ = x³ + y³ + 3x²y + 3xy²

Here x = p and y = 1

⇒ (p + 1)³ = p³ + 1³ + 3(p)²(1) + 3(p)(1)²

= p³ + 1³ + 3p² + 3p

Therefore (p + 1)³ = p³ + 1³ + 3p² + 3p

Using identity (a – b)³ = a³ - b³ – 3a²b + 3ab²

Here a = x and b = y

⇒ = x³ - – 3(x)²() + 3(x)

⇒ = x³ - – 2x²y + xy²

Therefore = x³ - – 2x²y + xy²

25x² + 16y² + 4z² - 40xy + 16yz - 20xz

25x² + 16y² + 4z² - 40xy + 16yz - 20xz = 25x² + 16y² + 4z² + (-40xy) + 16yz + (-20xz)

25x² can be written as (-5x)²

16y² can be written as (4y)²

4z² can be written as (2z)²

-40xy can be written as 2(-5x)(4y)

16yz can be written as 2(4y)(2z)

-20xz can be written as 2(-5x)(2z)

⇒ 25x² + 16y² + 4z² - 40xy + 16yz - 20xz = (-5x)² + (4y)² +

(2z)² + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) …(i)

Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Comparing (-5x)² + (4y)² + (2z)² + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) with a² + b² + c² + 2ab + 2bc + 2ca we get

a = -5x, b = 4y and c = 2z

therefore

(-5x)² + (4y)² + (2z)² + 2(-5x)(4y) + 2(4y)(2z) + 2(-5x)(2z) = (- 5x + 4y + 2z)²

From (i)

25x² + 16y² + 4z² - 40xy + 16yz - 20xz = (- 5x + 4y + 2z)²

9a² + 4b² + 16c² + 12ab - 16bc - 24ca

9a² + 4b² + 16c² + 12ab - 16bc - 24ca = 9a² + 4b² + 16c² + 12ab + (-16bc) + (-24ca)

9a² can be written as (3a)²

4b² can be written as (2b)²

16c² can be written as (-4c)²

12ab can be written as 2(3a)(2b)

-16bc can be written as 2(2b)(-4c)

-24ca can be written as 2(-4c)(3a)

⇒ 9a² + 4b² + 16c² + 12ab - 16bc - 24ca = (3a)² + (2b)² + (-4c)² + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) …(i)

Using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Comparing (3a)² + (2b)² + (-4c)² + 2(3a)(2b) + 2(2b)(-4c) + 2(-4c)(3a) with x² + y² + z² + 2xy + 2yz + 2zx we get

x = 3a, y = 2b and z = -4c

therefore

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

From (i)

9a² + 4b² + 16c² + 12ab - 16bc - 24ca = (3a + 2b – 4c)²