(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 (x);
(aii) 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;
the zeros of g (x).
(b) Find the third term when (\(\frac{x}{2}-1\))\(^8\)is expanded in descending powers of \(x\).
(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 = \(28\times \frac{x^6}{64}=\frac{7x^6}{16}\)
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