Ch – 2 Polynomials

Q1. Write (i) a binomial of degree 100
(ii) a monomial of degree 23
Q2. Write the coefficient of
i) x3 in 2x + x2 – 5x3 + x4
ii) x2 in (π/3) x2 + 7x – 3
iii) x in √3 - 2√2x + 4x2
Q3. If p(y) = 4 + 3y – y2 + 5y3, find (i) p(0) ii) p(2) iii) p(-1)
Q4. Find the zero of the polynomial
i) h(x) = 6x-1 ii) p(t) = 2t-3
Q5. Find the remainder in each case if p(x) is divided by g(x):
(i) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x; (ii) p(x) = 2x3 – 5x2 + 4x – 3, g(x) = 3x + 1.
Q6. In each of the following, find the value of k such that g(x) a factor of p(x):
(i) p(x) = 2x3 + kx2 + 11x + k + 3, g(x) = 2x – 1; (ii) p(x) = 25x4 + 10x3 – 5kx2 + 15kx – 1, g(x) = 2 + 5x.
Q7 . Find the value of k if p(x) = 3x3 – 2kx2 + 5x – 4 and q(x) = 2x3 – kx2 – 5k leave the same remainder when divided by x + 1.

The answer:
Q1: a binomial of degree 100:
for example: x^100 + 2 (it is called binomial because the number of term is 2: x^100 and 2)
Q2: the coefficients:
the main rule is:
if we have P(x)= a1X^p + a2X^p-1+ .... + a0 as polynomial, so the coefficients are (a1, a2, ... ao)
i) x3 in 2x + x2 – 5x3 + x4, the coefficient is -5
ii) x2 in (π/3) x2 + 7x – 3, the coefficient is (π/3)
iii) x in √3 - 2√2x + 4x2, the coefficient is - 2√2

Q3. If p(y) = 4 + 3y – y2 + 5y3, (i) p(0) can be found with
(i) p(0) = 4 + 3*0 – 0² + 5*0^3= 4, p(0)= 4, the same method to p(2)
ii) p(2)=4 + 3*2– 2² + 5*2^3=49, p(2)=49
iii) p(-1)= 4 + 3*(-1) – (-1)² + 5*(-1)^3 = -5, p(-1)= -5

Q4. the zero of the polynomial
i) h(x) = 6x-1, to find it, just egalize h(x) to 0, 6x-1 =0, it implies 6x=1, and x=1/6, so the zero is x=1/6
ii) p(t) = 2t-3, ii) p(t) = 2t-3=0 implies 2t=3 and the zero is t=3/2

the remainder in each case if p(x) is divided by g(x)
(i) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x

x3 – 6x2 + 2x – 4
R(x) = ------------------------ = (-x²/2 + 11x/4 + 3/8) - 35 / 8(-2x + 1)
1 – 2x

so the remainder is - 35 / 8

(ii) p(x) = 2x3 – 5x2 + 4x – 3, g(x) = 3x + 1.

2x3 – 5x2 + 4x – 3
R(x) = -------------------------- = 2x²/3 -17x /9 +53/27 - 134 /27(3x+1)
3x + 1

the remainder is - 134 /27

dominiquedhunter dominiquedhunter · Answer 2 · 2019-06-12T15:17:29+02:00

Answer:The answer:

Q1: a binomial of degree 100:

for example: x^100 + 2 (it is called binomial because the number of term is 2: x^100 and 2)

Q2: the coefficients:

the main rule is:

if we have P(x)= a1X^p + a2X^p-1+ .... + a0 as polynomial, so the coefficients are (a1, a2, ... ao)

i) x3 in 2x + x2 – 5x3 + x4, the coefficient is -5

ii) x2 in (π/3) x2 + 7x – 3, the coefficient is (π/3)

iii) x in √3 - 2√2x + 4x2, the coefficient is - 2√2

Q3. If p(y) = 4 + 3y – y2 + 5y3, (i) p(0) can be found with

(i) p(0) = 4 + 3*0 – 0² + 5*0^3= 4, p(0)= 4, the same method to p(2)

ii) p(2)=4 + 3*2– 2² + 5*2^3=49, p(2)=49

iii) p(-1)= 4 + 3*(-1) – (-1)² + 5*(-1)^3 = -5, p(-1)= -5

Q4. the zero of the polynomial

i) h(x) = 6x-1, to find it, just egalize h(x) to 0, 6x-1 =0, it implies 6x=1, and x=1/6, so the zero is x=1/6

ii) p(t) = 2t-3, ii) p(t) = 2t-3=0 implies 2t=3 and the zero is t=3/2

the remainder in each case if p(x) is divided by g(x)

(i) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x

x3 – 6x2 + 2x – 4

R(x) = ------------------------ = (-x²/2 + 11x/4 + 3/8) - 35 / 8(-2x + 1)

1 – 2x

so the remainder is - 35 / 8

(ii) p(x) = 2x3 – 5x2 + 4x – 3, g(x) = 3x + 1.

2x3 – 5x2 + 4x – 3

R(x) = -------------------------- = 2x²/3 -17x /9 +53/27 - 134 /27(3x+1)

3x + 1

the remainder is - 134 /27

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