[3 marks sum]

Question 13 Marks

Find the values of a and b when the polynomials $f(x)= 2x^2 -5x +a$ and $g(x)= 2x^2 + 5x +b$ both have a factor $(2x+1)$.

Answer

$(2 x+1) \Rightarrow x=-\frac{1}{2}$
Solving equation $(i),$ we get
$f \left(-\frac{1}{2}\right)=2 \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)-5 \times\left(-\frac{1}{2}\right)+ a =0 $
$\Rightarrow \frac{1}{2}+\frac{5}{2}+ a =0 $
$\Rightarrow a =-3 $
$g \left(-\frac{1}{2}\right)=2 \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)+5 \times\left(-\frac{1}{2}\right)+ b =0 $
$\Rightarrow \frac{1}{2}-\frac{5}{2}+ b =0 $
$\Rightarrow b =2 $
$\Rightarrow a =-3, b =2$

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Question 23 Marks

Prove by factor theorem that
$(2x - 1)$ is a factor of $6x^3 - x^2 - 5x +2$

Answer

$(2 x-1)$ is a factor of $6 x^3-x^2-5 x+2$
$2 x-1=0 \Rightarrow x=\frac{1}{2}$
Substituting this value, we get
$f \left(\frac{1}{2}\right)=6 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{1}{2}-5 \times \frac{1}{2}+2$
$=\frac{3}{4}-\frac{1}{4}-\frac{5}{2}+2$
$=\frac{1}{2}-\frac{5}{2}+2=-2+2=0$
Hence $(2 x-1)$ is a factor of $6 x^3-x^2-5 x+2$

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Question 33 Marks

Prove by factor theorem that
$(3x-2)$ is a factor of $18x^3 - 3x^2 + 6x -12$

Answer

$(3 x-2)$ is a factor of $18 x^3-3 x^2+6 x-12$
$3 x-2=0 \Rightarrow x=\frac{2}{3}$
Substituting this value, we get
$f \left(\frac{2}{3}\right)=18 \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right)-3 \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right)+6 \times\left(\frac{2}{3}\right)-8 $
$=\frac{16}{3}-\frac{4}{3}+4-8$
$=4+4-8$
$=0$
Hence $(3 x-2)$ Is a factor of $18 x^3-3 x^2+6 x-8$.

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Question 43 Marks

Prove by factor theorem that
$(2x+1)$ is a factor of $4x^3 + 12x^2 + 7x +1$

Answer

$(2 x+1)$ is a factor of $4 x^3+12 x^2+7 x+1$
$
2 x+1 \Rightarrow x=\frac{1}{2}
$
Substituting this value, we get
$
f \left(-\frac{1}{2}\right)=4 \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)+12 \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)+7 \times\left(-\frac{1}{2}\right)+1=0
$
Hence $(2 x+1) 1$ s a factor of $4 x^3+12 x^2+7 x+1$

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Question 53 Marks

What number should be added to polynomial $f(x)= 12x^3 + 16x^2 - 5x - 8$ so that the resulting polynomial is exactly divisible by $(2x - 1) ?$

Answer

$(2 x-1) \Rightarrow x=\frac{1}{2}$
When we substirute this value in the polynomial, whatever we get as a remainder (say a) should be added so that polynomial is exactly subtracted by the factor.
$f \left(\frac{1}{2}\right)=12 \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)+16 \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)-5 \times\left(\frac{1}{2}\right)-8+ a =0 $
$\Rightarrow \frac{3}{2}+4-\frac{5}{2}-8+ a =0$
$\Rightarrow a =5$
$+4-8+ a =\text { Hence answer }=5$

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Question 63 Marks

What number should be subtracted from the polynomial $f(x)= 2x^3 - 5x^2 +8x -17$ so that the resulting polynomial is exactly divisible by $(2x - 5)?$

Answer

$(2 x-5)=0 \Rightarrow x=\frac{5}{2}$
When we substitute this value in the polynomial, whatever we get as a remainder (say a) should be subtracted so that polynomial is exactly subtracted by the factor.
$f \left(\frac{5}{2}\right)=2 \times\left(\frac{5}{2}\right) \times\left(\frac{5}{2}\right) \times\left(\frac{5}{2}\right)-5 \times\left(\frac{5}{2}\right) \times\left(\frac{5}{2}\right)+8 \times\left(\frac{5}{2}\right)-17- a =0 $
$\Rightarrow \frac{125}{4}-\frac{125}{4}+20-17- a =0 $
$\Rightarrow a =3$
Hence answer $=3$

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Mathematics [3 marks sum]

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