(ai) A quadratic polynomial, g (x) has (2x + 1) as a factor. If g (x) is divided by (x - 1) and (x - 2), the remainder are -6 and -5 respectively. Find; g (...

(ai) Let the quadratic equation be \(ax^2+bx+c\)

\(g(x)=ax^2+bx+c\)

Since \((2x+1)\) is a factor,\(x=-\frac{1}{2}\)is a root

\(∴g(-\frac{1}{2})=a(-\frac{1}{2})^2+b(-\frac{1}{2})+c=0\)

\(=a(\frac{1}{4})-b(\frac{1}{2})+c=0\)

Multiply through by 4

=a-2b+4c=0-----(i)

For(x-1)and(x-2),x=1 and x = 2 respectively

So,

\(g(1)=a(1)^2+b(1)+c=-6\)

=a+b+c=-6-----(ii)

and

\(g(2)=a(2)^2+b(2)+c=-5\)

= 4a+2b+c=-5

Adding equations (i) and (iii) gives;

=5a+5b=-5

Divide through by 5

=a+b=-1......(iv)

Adding equation (i) to 2 times equation (iv) gives:

3a+6c=-12.....(v)

Equation (v) minus three times equation (iv) gives:

=3c=-12-(-3)

=3c=-9

=c\(\frac{-9}{3}=-3\)

Substitute (-3) for c in equation (v)

=3a+6(-3)=-12

=3a-18=-12

=3a=-12+18

=3a=6

=a=\(\frac{6}{3}=2\)

Substitute 2 for a and (-3) for c in equation (ii)

=2+6-3=-6

=b-1=-6

=b=-6+1

=b=-5

\(\therefore g(x)=2x^2-5x-3\)

(aii) \(2\times2 - 5x - 3\)

\(= 2\times2 - 6x + x - 3\)

\(= 2x(x - 3) + 1(x - 3)\)

\(= (x - 3)(2x + 1)\)

zeros of \(g(x)\) are \(2x + 1 = 0\) and \(x - 3 = 0\)

∴ zeros of \(g(x)\) are -\frac{1}{2}\) and \(3\)

(b) \((\frac{x}{2}-1)^8\)

rth term is given as \(^nC_r-1 a^n-(r-1)b^r-1\)

\(a=\frac{x}{2},b=-1,n=8,r=3,r-1=2\)

3rd term = \(^8C_2(\frac{x}{2})^8-2(-1)^2\)

= \(28×(\frac{x}{2})^6\times1\)

∴3rd term = \(28\times \frac{x^6}{64}=\frac{7x^6}{16}\)