SEE Mathematics: Polynomial

1. If 2x³ – 4x² + kx + 10 is divided by (x + 2), the remainder is 4. Find the value of k using remainder theorem.
Solution:
Here,
     Let  p(x) = 2x³ – 4x² + kx + 10, 
       d(x) = x + 2 and R = 4 .
Comparing x + 2 with x - a , we get
        a = -2 
Now, Using remainder theorem,
     p(-2)  = 2 × (-2)³ – 4 × (-2)² + k(-2) + 10
⇒ 4 = - 16 – 16 – 2k + 10 
⇒ 26 = -2k 
∴    k = - 13 .
Thus, the value of k is -13. 

2. If x – 3 is a factor of 2x³ – x² + 10x – k, then find the value of k.
Solution:
Here,
p(x) = 2x³ – x² + 10x – k and a factor = (x – 3).
Comparing x - 3 with  x - a, we get
 a = 3 .
Now,
Using factor  theorem.
    p(3) = 0
⇒ 2  × 3³ – 3² + 10  × 3 – k
⇒ 0 = 54 – 9 + 30 – k
∴    k = 75.
Thus, the value of k is 75. 

3. If x – 5 is a factor of x³ + px² + 4x + 5, find the value of p.
Solution:
Here,

p(x) = x³ + px² + 4x + 5 and a factor = (x – 5)

Comparing x - 5 with x - a, we get

x = 5 . 

Now,

Using factor theorem, 

              p(5) =  0

⇒  5³ + p × 5² + 4 × 5 + 5=0

⇒  125 + 25p + 25= 0

⇒   25p = -150

∴     p = - 6

Thus, the value of p is -6.

4. Using remainder theorem, find the remainder when 8x³ – 4x² + 2x – 5 is divided by 2x – 1.

Solution:

Here,

p(x) = 8x³ – 4x² + 2x – 5 and d(x) = 2x – 1= 2( x - 1/2)

Comparing x - 1/2 with x- a, we get,

a = 1/2. 

Now,Using remainder theorem   

  R = p(½)

⇒R = 8×(½)³ - 4(½)² + 2(½) - 5

⇒ R = 8 × 1 - 4 × 1 + 1 – 5

⇒ R = 1 – 1 + 1 – 5

∴   R = - 4.

Thus, the remainder is -4.

5. If x – 5 is a factor of 2x³ – 7px + (p – 12), find the value of p.

Solution:

Here,

p(x) = 2x³ – 7px + (p – 12) and factor = x – 5.

 Comparing x - 5 with x - a, we get, a = 5.

 Now,By using factor theorem

 p(5) = 0

 ⇒ 2 × 5³ – 7p × 5 + p – 12 = 0

 ⇒ 250 – 35p + p – 12  = 0 

 ⇒ 34p = 238 

  ∴    p = 7 .

 Thus, the value of p is 7.

6. If x³ – 19x – 30 = (x + 2). Q(x), find Q(x) by using synthetic division method.

Solution:
Here,
x³ – 19x – 30 = (x + 2). Q(x) 
Comparing (x + 2) with (x – a), then a = -2. 
Using synthetic division, 


∴ Q(x) = x² – 2x – 15.

7. State factor theorem. Use factor theorem to determine whether x + 3 is a factor of the polynomial x³ – 8x + 3. 
Solution:
i)
Statement of factor theorem:
"If a polynomial p(x) is divided by (x – a) and f(a) = R = 0 then (x – a) is a factor of p(x).
ii)
Here, p(x) = x³ – 8x + 3 and d(x) = x + 3
Comparing (x + 3) with (x – a) then, a = -3
Now,
        P(a) = P(3)
                 = (-3)³ – 8 ×  (-3) + 3 
                 = -27 + 24 + 3
                 = 0 .
Since P(3) = 0 shows that (x + 3)  is a factor  of P(x). 

 8. If x³ – 21x – 20 = (x + 1). Q(x), find Q(x) by using synthetic division method.
Solution: 
 Here,
x³ – 19x – 30 = (x + 2). Q(x) 
Comparing (x - a) with (x + 1), then a = -1. 
Using synthetic division, 

∴ Q(x) = x² – x – 20.

9. If the polynomial x² + 6x² + kx + 10 is divided by (x + 2), the remainder is 4, find the value of k using the remainder theorem.
Solution:
Here, p(x) = x³ + 6x² + kx + 10, 
d(x) = x + 2 and R = 4 
Using remainder theorem,
     p(-2) =4
⇒ (-2)³ + 6 × (-2)²+ k (-2) + 10 = 4.
⇒ 2k = 22 
⇒ 2k = 22 
∴   k = 11.
 
Thus, the value of k is 11. 
 
10. If (x + 2) is a factor of x³ – 19x – p, find the value of p.
Solution:
Here,
p(x) = x³ – 19x – p and x + 2 is factor of p(x),
Comparing x + 2 with x - a, we get  a  = - 2.
By factor theorem,
So, p(-2) = 0 
⇒(-2)³ – 19 (-2) – p = 0
⇒ -8 + 38 – p = 0
∴   p = 30 .
Thus, the value of p is 30. 

 11. Solve: 3x³ = 7x² – 4
Solution:
Here,
     3x³ = 7x² – 4
i.e.  3x³ – 7x² + 0.x + 4 = 0.
Let f(x)=3x³ – 7x² + 0.x + 4 = 0
Possible roots of f(x) are ±1,±2,±3,±4,±6,±12
For, x = 1
f(1)= 3.1³ - 7.1² + 4
       = 0
∴ 1 is a root of f(x).
Dividing f(x) by (x-1) by synthetic division method:
 = R.

∴ (x – 1) (3x² – 4x – 4) = 3x³ – 7x² + 0.x + 4
i.e. (x – 1) (3x² – 4x – 4) = 0
⇒ (x – 1) (3x² – 6x + 2x – 4) = 0
⇒ (x – 1) {3x(x – 2) + 2(x – 2)} = 0
⇒ (x – 1) (x – 2) (3x + 2) = 0
⇒ Either, x – 1 = 0 i.e. x = 1
Or, x – 2 = 0 i.e. x = 2
Or, 3x + 1 = 0 i.e. x = - 2/3.
Thus, x = 1 or 2 or - 23 is the solution.

13. Solve: y3 – 6y² + 11y – 6 = 0
Solution:
Here, 
Let f(y) y³ – 6y² + 11y – 6 = 0     
Possible factors of f(x) are ±1,±2,±3,±6
For y = 1, 
f(1)= 1³ - 6.1² + 11.1 - 6
       = 0
∴ 1 is a root of f(x).
Dividing f(y) by (y-1) by synthetic division method:


∴ (y – 1) (y² – 5y + 6) = y³ – 6y² + 11y – 6
i.e. y³ – 6y² + 11y – 6 = 0 
⇒ (y – 1) (y² – 5y + 6) = 0 
⇒ (y – 1) (y² – 3y – 2y + 6) = 0 
⇒ (y – 1) {y(y – 3) - 2(y – 3)} = 0 
⇒ (y – 1) (y – 3) (y - 2) = 0 
⇒  Either, y – 1 = 0 i.e. y=1
Or, y – 3 = 0 i.e. y=3
Or, y - 2 = 0 i.e. y=2
Thus, y = 1 or 3 or 2 is the solutions of the given equation. 

Exercise
14. Solve: 6x³ + x² – 19x + 6 = 0
15. Solve: 2x³ + 6 - 3x² – 11x = 0
16. Solve: x³ - 3x² – 10x + 24 = 0  
17. Solve: 6x³ = 4 - 13x²
18. Solve: 2x³ + 3x² – 11x - 6 = 0
19. Solve: x³ = 7x² - 36
20. Solve: x³ -21x – 20 = 0 
21. Solve: x³ - 4x² + x + 6 = 0