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
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

At a constant it will take one hour to travel 32 mi.at that speed it will take 1/2 hr to travel the remaining 16 mi.it will take 1+1/2 hr to travel the 48mi @ 32mph.so 1 hr = 60 min& 1/2 hr = 30 min1+1/2 hr = 90 min.there are 60s in each 90 × 60 = 3600 seconds.the real question seriously is he why are they driving so slow? ! lol.