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RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

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Q1. If f(x) = 2x3–13x2+17x+12 , Find

1. f(2)
2. f(-3)
3. f(0)

Sol :

The given polynomial is f(x) = 2x3–13x2+17x+12

1. f(2)

we need to substitute the ‘2‘ in f(x)

f(2) = 2(2)3–13(2)2+17(2)+12

= (2 * 8) – (13 * 4) + (17 * 2) + 12

= 16 – 52 + 34 + 12

= 10

therefore f(2) = 10

2. f(-3)

we need to substitute the ‘ (-3) ‘ in f(x)

f(-3) =2(−3)3–13(−3)2+17(−3)+12

= (2 * -27) – ( 13 * 9) – ( 17 * 3) + 12

= -54 – 117 – 51 + 12

= -210

therefore f(-3) = -210

3. f(0)

we need to substitute the ‘(0)‘ in f(x)

f(0) = 2(0)3–13(0)2+17(0)+12

= ( 2 * 0) – ( 13 * 0) + (17 * 0) + 12

= 0 – 0 + 0 + 12

= 12

therefore f(0) = 12


Q2. Verify whether the indicated numbers are zeros of the polynomial corresponding to them in the following cases :

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9
RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

Sol :

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

we know that ,

f(x) = 3x + 1
RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

= -1 + 1

= 0

Since, the result is 0 RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9 is the root of 3x + 1

(2) f(x) = x2–1,x = (1,−1)

we know that,

f(x) = x2 – 1

Given that x = (1 , -1)

substitute x = 1 in f(x)

f(1) = 12 – 1

= 1 – 1

= 0

Now , substitute x = (-1) in f(x)

f(-1) = (−1)2 – 1

= 1 – 1

= 0

Since , the results when x = (1 , -1) are 0 they are the roots of the polynomial f(x) = x2 – 1

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

Sol :

We know that

g(x) = 3x2–2

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

= 4 – 2

= 2 ≠ 0
Now, Substitute RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9
RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

Since, the results when RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9 are not 0, they are roots of 3x2–2

(4) p(x) = x3–6x2+11x–6 , x = 1, 2, 3

Sol :

We know that ,

p(x) = x3–6x2+11x–6

given that the values of x are 1, 2 , 3

substitute x = 1 in p(x)

p(1) = 13–6(1)2+11(1)–6

= 1 – (6 * 1) + 11 – 6

= 1 – 6 + 11 – 6

= 0

Now, substitute x = 2 in p(x)

P(2) =23–6(2)2+11(2)–6

= (2 * 3) – ( 6 * 4) + ( 11 * 2) – 6

= 8 – 24 – 22 – 6

= 0

Now, substitute x = 3 in p(x)

P(3) = 33–6(3)2+11(3)–6

= ( 3 * 3) – (6 * 9) + (11 * 3) – 6

= 27 – 54 + 33 – 6

= 0

Since , the result is 0 for x = 1, 2, 3 these are the roots of x3–6x2+11x–6

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

we know that ,

f(x) = 5x–π

Given that , x = 4/5

Substitute the value of x in f(x)

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

= 4 – π

≠ 0

Since , the result is not equal to zero ,  x = 4/5 is not the root of the polynomial 5x – π

(6) f(x) = x2 , x = 0

Sol :

we know that , f(x) = x2

Given that value of x is ‘ 0 ’

Substitute the value of x in f(x)

f(0) = 02

= 0

Since, the result is zero , x = 0 is the root of x2

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

Sol :

We know that,

f(x) = lx + m

Given , that   RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

Substitute the value of x in f(x)

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

= -m + m

= 0  RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9  is the root of lx + m

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

Sol :

We know that ,

f(x) = 2x + 1

Given that x = 1/2

Substitute the value of x and f(x)

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

= 1 + 1

= 2 ≠ 0

Since , the result is not equal to zero

x = 1/2 is the root of 2x + 1


Q3. If x = 2 is a root of the polynomial f(x) = 2x2–3x+7a, Find the value of a

Sol :

We know that , f(x) = 2x2–3x+7a

Given that x = 2 is the root of f(x)

Substitute the value of x in f(x)

f(2) = 2(2)2–3(2)+7a

= (2 * 4) – 6 + 7a

= 8 – 6 + 7a

= 7a + 2

Now, equate 7a + 2 to zero

⇒ 7a + 2 = 0

⇒ 7a = -2

⇒  RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

The value of RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9


Q4. If  RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9 is zero of the polynomial p(x) = 8x3–ax2–x+2 , Find the value of a

Sol :

We know that , p(x) = 8x3–ax2–x+2

Given that the value of  RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

Substitute the value of x in f(x)

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9
RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

To , find the value of a , equate  RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9 to zero

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

On taking L.C.M

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9  

⇒ 6 – a = 0

⇒ a = 6


Q5. If x = 0 and x = -1 are the roots of the polynomial f(x) = 2x3–3x2+ax+b , Find the of a and b.

Sol :

We know that ,  f(x) = 2x3–3x2+ax+b

Given , the values of x are 0 and -1

Substitute x = 0 in f(x)

f(0) = 2(0)3–3(0)2+a(0)+b

= 0 – 0 + 0 + b

= b     ——— 1

Substitute x = (-1) in f(x)

f(-1) = 2(−1)3–3(−1)2+a(−1)+b

= -2 – 3 – a + b

= -5 – a + b    ———— 2

We need to equate equations 1 and 2 to zero

b = 0 and -5 – a + b = 0

since, the value of b is zero

substitute  b = 0 in equation 2

⇒ -5 – a = -b

⇒ -5 – a = 0

a = -5

the values of a and b are -5 and 0 respectively


Q6. Find the integral roots of the polynomial f(x) = x3+6x2+11x+6

Sol :

Given , that f(x) = x3+6x2+11x+6

Clearly we can say that, the polynomial f(x) with an integer coefficient and the highest degree term coefficient which is known as leading factor is 1.

So, the roots of f(x) are limited to integer factor of 6, they are

±1, ±2, ±3, ±6

Let x = -1

f(-1) = (−1)3+6(−1)2+11(−1)+6

= -1 + 6 -11 + 6

= 0

Let x = -2

f(-2) = (−2)3+6(−2)2+11(−2)+6

= -8 – (6 * 4) – 22 + 6

= -8 + 24 – 22 + 6

= 0

Let x = -3

f(-3) = (−3)3+6(−3)2+11(−3)+6

= -27 – (6 * 9) – 33 + 6

= -27 + 54 – 33 + 6

= 0

But from all the given factors only -1 , -2 , -3 gives the result as zero .

So, the integral multiples of x3+6x2+11x+6 are -1 , -2 , -3


Q7. Find the rational roots of the polynomial f(x) = 2x3+x2–7x–6

Sol :

Given that f(x) = 2x3+x2–7x–6

f(x) is a cubic polynomial with an integer coefficient . If the rational root in the form of p/q , the values of p are limited to factors of 6 which are ±1, ±2, ±3, ±6

and the values of q are limited to the highest degree coefficient i.e 2 which are ±1, ±2

here, the possible rational roots are

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

Let , x = -1

f(-1) = 2(−1)3+(−1)2–7(−1)–6

= -2 + 1 +7 – 6

= -8 + 8

= 0

Let , x = 2

f(-2) = 2(2)3+(2)2–7(2)–6

= ( 2 * 8) + 4 -14 – 6

= 16 + 4 -14 – 6

= 20 – 20

= 0

RD Sharma Solutions Ex-6.2, Factorization Of Polynomials, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

But from all the  factors only -1 , 2 and −3/2 gives the result as zero

So, the rational roots of 2x3+x2–7x–6 are -1, 2 and −3/2

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