Q10. y3–7y+6
Sol :
Given, f(y) = y3–7y+6
The constant term in f(y) is 6
The factors are ±1, ±2, ±3, ±6
Let , y – 1 = 0
⇒ y = 1
f(1) = (1)3–7(1)+6
= 1 – 7 + 6
= 0
So, (y – 1) is the factor of f(y)
Similarly, (y – 2) and (y + 3) are also the factors
Since, f(y) is a polynomial which has degree 3, it cannot have more than 3 linear factors
⇒ f(y) = k(y – 1)( y – 2)(y + 3)
⇒ y3–7y+6 = k(y – 1)( y – 2)(y + 3) ———– 1
Substitute k = 0 in eq 1
⇒ 0 – 0 + 6 = k(-1)(-2)(3)
⇒ 6 = 6k
⇒ k = 1
y3–7y+6 = (1)(y – 1)( y – 2)(y + 3)
y3–7y+6 = (y – 1)( y – 2)(y + 3)
Hence, y3–7y+6 = (y – 1)( y – 2)(y + 3)
Q11. x3–10x2–53x–42
Sol :
Given , f(x) = x3–10x2–53x–42
The constant in f(x) is -42
The factors of -42 are ±1, ±2, ±3, ±6, ±7, ±14, ±21,±42
Let, x + 1 = 0
⇒ x = -1
f(-1) = (−1)3–10(−1)2–53(−1)–42
= -1 – 10 + 53 – 42
= 0
So., (x + 1) is the factor of f(x)
Now, divide f(x) with (x + 1) to get other factors
By long division,
x2 – 11x – 42
x + 1 x3 – 10x2 – 53x – 42
x3 + x2
(-) (-)
-11x2 – 53x
-11x2 – 11x
(+) (+)
-42x – 42
-42x – 42
(+) (+)
0
⇒ x3 – 10x2 – 53x – 42 = (x + 1) (x2 – 11x – 42)
Now,
x2 – 11x – 42 = x2 – 14x + 3x – 42
= x(x – 14) + 3(x – 14)
= (x + 3)(x – 14)
Hence, x3 – 10x2 – 53x – 42 = (x + 1) (x + 3)(x – 14)
Q12. y3–2y2–29y–42
Sol :
Given, f(x) = y3–2y2–29y–42
The constant in f(x) is -42
The factors of -42 are ±1, ±2, ±3, ±6, ±7, ±14, ±21,±42
Let, y + 2 = 0
⇒ y = -2
f(-2) = (−2)3–2(−2)2–29(−2)–42
= -8 -8 + 58 – 42
= 0
So, ( y + 2) is the factor of f(y)
Now, divide f(y) with (y + 2) to get other factors
By, long division
y2 – 4y – 21
y + 2 y3 – 2y2 – 29y – 42
y3 + 2y2
(-) (-)
-4y2 – 29y
-4y2 – 8y
(+) (+)
-21y – 42
-21y – 42
(+) (+)
0
⇒ y3 – 2y2 – 29y – 42 = (y + 2) (y2 – 4y – 21)
Now,
y2 – 4y – 21 = y2 – 7y + 3y – 21
= y(y – 7) +3(y – 7)
= (y – 7)(y + 3)
Hence, y3 – 2y2 – 29y – 42 = (y + 2) (y – 7)(y + 3)
Q13. 2y3–5y2–19y+42
Sol :
Given, f(x) = 2y3–5y2–19y+42
The constant in f(x) is +42
The factors of 42 are ±1, ±2, ±3, ±6, ±7, ±14, ±21,±42
Let, y – 2 = 0
⇒ y = 2
f(2) = 2(2)3–5(2)2–19(2)+42
= 16 – 20 – 38 + 42
= 0
So, (y – 2) is the factor of f(y)
Now, divide f(y) with (y – 2) to get other factors
By, long division method
2y2 – y – 21
y – 2 2y3 – 5y2 -19y + 42
2y3 – 4y2
(-) (+)
-y2 – 19y
-y2 + 2y
(+) (-)
-21y + 42
-21y + 42
(+) (-)
0
⇒ 2y3 – 5y2 -19y + 42 = (y – 2) (2y2 – y – 21)
Now,
2y2 – y – 21
The factors are (y + 3) (2y – 7)
Hence, 2y3 – 5y2 -19y + 42 = (y – 2) (y + 3) (2y – 7)
Q14. x3+13x2+32x+20
Sol:
Given, f(x) = x3+13x2+32x+20
The constant in f(x) is 20
The factors of 20 are ±1, ±2, ±4, ±5, ±10, ±20
Let, x + 1 = 0
⇒ x = -1
f(-1) = (−1)3+13(−1)2+32(−1)+20
= -1 + 13 – 32 + 20
= 0
So, (x + 1) is the factor of f(x)
Divide f(x) with (x + 1) to get other factors
By, long division
x2 + 12x + 20
x + 1 x3 + 13x2 +32x + 20
x3 + x2
(-) (-)
12x2 + 32x
12x2 + 12x
(-) (-)
20x – 20
20x – 20
(-) (-)
0
⇒ x3 + 13x2 +32x + 20 = (x + 1)( x2 + 12x + 20)
Now,
x2 + 12x + 20 = x2 + 10x + 2x + 20
= x(x + 10) + 2(x + 10)
The factors are (x + 10) and (x + 2)
Hence, x3 + 13x2 +32x + 20 = (x + 1)(x + 10)(x + 2)
Q15. x3–3x2–9x–5
Sol :
Given, f(x) = x3–3x2–9x–5
The constant in f(x) is -5
The factors of -5 are ±1, ±5
Let, x + 1 = 0
⇒ x = -1
f(-1) = (−1)3–3(−1)2–9(−1)–5
= -1 – 3 + 9 – 5
= 0
So, (x + 1) is the factor of f(x)
Divide f(x) with (x + 1) to get other factors
By, long division
x2 – 4x – 5
x + 1 x3 – 3x2 – 9x – 5
x3 + x2
(-) (-)
-4x2 – 9x
-4x2 – 4x
(+) (+)
-5x – 5
-5x – 5
(+) (+)
0
⇒ x3 – 3x2 – 9x – 5 = (x + 1)( x2 – 4x – 5)
Now,
x2 – 4x – 5 = x2 – 5x + x – 5
= x(x – 5) + 1(x – 5)
The factors are (x – 5) and (x + 1)
Hence, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x + 1)
Q16. 2y3+y2–2y–1
Sol :
Given , f(y) = 2y3+y2–2y–1
The constant term is 2
The factors of 2 are ±1, ± 1/2
Let, y – 1= 0
⇒ y = 1
f(1) = 2(1)3+(1)2–2(1)–1
= 2 + 1 – 2 – 1
= 0
So, (y – 1) is the factor of f(y)
Divide f(y) with (y – 1) to get other factors
By, long division
2y2 + 3y + 1
y – 1 2y3 + y2 – 2y – 1
2y3 – 2y2
(-) (+)
3y2 – 2y
3y2 – 3y
(-) (+)
y – 1
y – 1
(-) (+)
0
⇒ 2y3 + y2 – 2y – 1 = (y – 1) (2y2 + 3y + 1)
Now,
2y2 + 3y + 1 = 2y2 + 2y + y + 1
= 2y(y + 1) + 1(y + 1)
= (2y + 1) ( y + 1) are the factors
Hence, 2y3 + y2 – 2y – 1 = (y – 1) (2y + 1) ( y + 1)
Q17. x3–2x2–x+2
Sol :
Let, f(x) = x3–2x2–x+2
The constant term is 2
The factors of 2 are ±1, ±12
Let, x – 1= 0
⇒ x = 1
f(1) = (1)3–2(1)2–(1)+2
= 1 – 2 – 1 + 2
= 0
So, (x – 1) is the factor of f(x)
Divide f(x) with (x – 1) to get other factors
By, long division
x2 – x – 2
x – 1 x3 – 2x2 – y + 2
x3 – x2
(-) (+)
-x2 – x
-x2 + x
(+) (-)
– 2x + 2
– 2x + 2
(+) (-)
0
⇒ x3 – 2x2 – y + 2 = (x – 1) (x2 – x – 2)
Now,
x2 – x – 2 = x2 – 2x + x – 2
= x(x – 2) + 1(x – 2)
=(x – 2)(x + 1) are the factors
Hence, x3 – 2x2 – y + 2 = (x – 1)(x + 1)(x – 2)
Q18. Factorize each of the following polynomials :
1. x3+13x2+31x–45 given that x + 9 is a factor
2. 4x3+20x2+33x+18 given that 2x + 3 is a factor
Sol :
1. x3+13x2+31x–45 given that x + 9 is a factor
let, f(x) = x3+13x2+31x–45
given that (x + 9) is the factor
divide f(x) with (x + 9) to get other factors
by , long division
x2 + 4x – 5
x + 9 x3 + 13x2 + 31x – 45
x3 + 9x2
(-) (-)
4x2 + 31x
4x2 + 36x
(-) (-)
-5x – 45
-5x – 45
(+) (+)
0
⇒ x3 + 13x2 + 31x – 45 = (x + 9)( x2 + 4x – 5)
Now,
x2 + 4x – 5 = x2 + 5x – x – 5
= x(x + 5) -1(x + 5)
= (x + 5) (x – 1) are the factors
Hence, x3 + 13x2 + 31x – 45 = (x + 9)(x + 5)(x – 1)
2. 4x3+20x2+33x+18 given that 2x + 3 is a factor
let, f(x) = 4x3+20x2+33x+18
given that 2x + 3 is a factor
divide f(x) with (2x + 3) to get other factors
by, long division
2x2 + 7x + 6
2x + 3 4x3 + 20x2 + 33x + 18
4x3 + 6x2
(-) (-)
14x2 – 33x
14x2 – 21x
(-) (+)
12x + 18
12x + 18
(-) (-)
0
⇒ 4x3 + 20x2 + 33x + 18 = (2x + 3) (2x2 + 7x + 6)
Now,
2x2 + 7x + 6 = 2x2 + 4x + 3x + 6
= 2x(x + 2) + 3(x + 2)
= (2x + 3)(x + 2) are the factors
Hence, 4x3 + 20x2 + 33x + 18 = (2x + 3)(2x + 3)(x + 2)
1. What is the concept of factorization of polynomials? |
2. How can we factorize a quadratic polynomial? |
3. Can all polynomials be factorized? |
4. What is the importance of factorization of polynomials? |
5. Can factorization of polynomials be applied in real-life situations? |
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