Question

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Answer 1

Answer:

5. quotient = x³+13x²-20x+28 ;reminder = -53

6. 2(-2)²-3(-2)(5k+4) = 0

8+6(5k+4) = 0

6(5k+4) = -8

(5k+4) = -8/6 = -4/3

5k = -4/3 -4 = -16/3

k = -16/15

7. summation of zeroes = 0

2k+5 = 0

k = -5/2

8. exactly divisible by x-2,

12-p = -4

p = 12+4 = 16

Answer 2

Step-by-step explanation:

5)

Given Polynomial is x^4 +15x^3 +6x^2 - 12x +3

Divisor = x+2

We know that

Remainder Theorem:-

Let P(x) be a polynomial of the degree is greater than or equal to 1 and (x-a) is another linear polynomial,If P(x) is divided by (x-a) then remainder is P(a).

So, P(x) is divided by (x+2) then the remainder is P(-2).

P(-2) =

=> (-2)^4+15(-2)^3+6(-2)^2-12(-6)+3

=> 16+15(-8)+6(4)+72+3

=16-120+24+72+3

=> 115-120

=> -5

P(-2) = -5

The remainder is -5

6)

Given Polynomial P(x) = 2x^2-3x(5k+4)

Zero = -2

We know that

If -2 is zero of the Polynomial then it satisfies the given Polynomial

=>P(-2) = 0

=> 2(-2)^2-3(-2)(5k+4)=0

=> 2(4)+6(5k+4) = 0

=> 8+30k+24 = 0

=> 30k+32 = 0

=> 30 k = -32

=> k = -32/30

=> k = -16/15

The value of k = -16/15

7)

Given quardratic polynomial is 3x^2+(2k-5)x-7

On Comparing this with the standard quadratic Polynomial ax^2+bx+c

a = 3

b=2k-5

c = -7

Given zeroes α and -α

sum of the zeroes α +(-α) = -b/a

=> α +(-α) = -(2k-5)/3

=> 0 = -(2k-5)/3

=> 0×3 = -(2k-5)

=> 0 = -(2k-5)

=> -2K+5 = 0

=> -2k = -5

=> 2k = 5

=> k = 5/2

The value of k = 5/2

8)

Given Polynomial P(x) = x^3+4x^2-px+8

Given that P(x) is exactly divisible by (x-2) then by

factor theorem it is a factor and P(2) = 0

=> (2)^3+4(2)^2-p(2)+8 = 0

=>8+4(4)-2p+8 = 0

=> 8+16-2p+8 = 0

=> 32-2p = 0

= >32 = 2p

=> 2p = 32

=> p = 32/2

=> p = 16

The value of p = 16