[5 marks sum]

Question 15 Marks

If $(x + 1)$ and $(x – 2)$ are factors of $x^3 + (a + 1)x^2 – (b – 2)x – 6$, find the values of a and b. And then, factorise the given expression completely.

Answer

Let $f(x)=x^3+(a+1) x^2-(b-2) x-6$
Since, (x + 1) is a factor of f(x).
Remainder $= f(-1) = 0$
$(-1)^3+(a+1)(-1)^2-(b-2)(-1)-6=0$
$-1 + (a + 1) + (b – 2) – 6 = 0$
$a + b – 8 = 0 …(i)$
Since, (x – 2) is a factor of f(x).
Remainder $= f(2) = 0$
$(2)^3+(a+1)(2)^2-(b-2)(2)-6=0$
$8 + 4a + 4 – 2b + 4 – 6 = 0$
$4a – 2b + 10 = 0$
$2a – b + 5 = 0 …(ii)$
Adding (i) and (ii), we get,
$3a – 3 = 0$
$a = 1$
Substituting the value of a in (i), we get,
$1 + b – 8 = 0$
$b = 7$
$\therefore f(x)=x^3+2 x^2-5 x-6$
Now, $(x + 1)$ and $(x – 2)$ are factors of f(x).
Hence, $(x + 1) (x – 2) = x^2 – x – 2$ is a factor of $f(x).$

$f(x)=x^3+2 x^2-5 x-6=(x+1)(x-2)(x+3)$

View full question & answer→

Question 25 Marks

$(3x + 5)$ is a factor of the polynomial $(a – 1)x^3 + (a + 1)x^2 – (2a + 1)x – 15$. Find the value of ‘a’, factorise the given polynomial completely.

Answer

Let $f(x)=(a-1) x^3+(a+1) x^2-(2 a+1) x-15$
it is given that (3x + 5) is a factor of f(x)
∴ Remainder = 0
$f\left(\frac{-5}{3}\right)=0$
$(a-1)\left(-\frac{5}{3}\right)^3+(a+1)\left(\frac{-5}{3}\right)^2-(2 a+1)\left(\frac{-5}{3}\right)-15=0$
$(a-1)\left(\frac{-125}{27}\right)+(a+1)(a+1)\left(\frac{25}{9}\right)-(2 a+1)\left(\frac{-5}{3}\right)-15=0$
$\frac{-125(a-1)+75(a+1)+45(2 a+1)-405}{27}=0$
$-125 a+125+75 a+75+90 a+45-405=0$
$40 a-160=0$
$40 a=160$
$a=4$
$\therefore f(x)=(a-1) x^3+(a+1) x^2-(2 a+1) x-15$
$=3 x^3+5 x^2-9 x-15$

$\therefore 3 x^3+5 x^2-9 x-15=(3 x+5)\left(x^2-3\right)$
$=(3 x+5)(x+\sqrt{3})(x-\sqrt{3})$

View full question & answer→

Question 35 Marks

The expression $4 x^3-b x^2+x-c$ leaves remainders 0 and 30 when divided by $x+1$ and $2 x-3$ respectively. Calculate the values of b and c. Hence, factorise the expression completely.

Answer

Let $f(x) = 4x^3- bx^2+ x-c$
it is given that when $f(x)$ is dividend by $(x+1)$, the remainder is 0
$\therefore f(-1)=0$
$4(-1)^3-b(1)^2+(-1)-c=0$
$-4-b-1-c=0$
$b+c+5=0 \ldots \ldots . .(1)$
It is given that when f(x) is divided by (2x-3) the remainder is 30.
$\therefore f\left(\frac{3}{2}\right)=30$
$4\left(\frac{3}{2}\right)^3-b\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)-c=30$
$\frac{27}{2}-\frac{9 b}{4}+\frac{3}{2}-c=30$
$54-9 b+6-4 c-120=0$
$9 b+4 c+60=0$ ................. (2)
Multiplying (1) by 4 and subtracting it from (2), we get,
$5 b+40=0$
$b=-8$
Substituting the value of b in (1), we get,
$c=-5+8=3$
Therefore, $f(x)=4x^3+8 x^{x-3}$
Now, for $x-1$, wr get,
$f(x)=f(-1)=4(-1)^3+8(-1)^2+(-1)-3=-4+8-1-3=0$
Hence, (x+1) is a factor of f (x)

$\therefore 4 x^3+8 x^2+x-3=(x+1)\left(4 x^2+4 x-3\right)$
$=(x+1)\left(4 x^2+6 x-2 x-3\right)$
$=(x+1)[2 x(2 x+3)-(2 x+3)]$
$=(x+1)(2 x+3)(2 x-1)$

View full question & answer→

Question 45 Marks

Using the Remainder Theorem, factorise each of the following completely.$3x^3+ 2x^2 − 19x + 6$

Answer

For $x=2,$ the value of the given
expression $3 x^3+2 x^2-19 x+6$
$=3(2)^3+2(2)^2-19(2)+6$
$=24+8-38+6$
$=0$
$\Rightarrow x-2$ is a factor of $3 x^3+2 x^2-19 x+6$
Now let us do long division

Thus we have ,
$3 x^3+2 x^2-19 x+6=(x-2)\left(3 x^2+8 x-3\right)$
$=(x-2)\left(3 x^2+9 x-x-3\right)$
$=(x-2)(3 x(x+3)-(x+3))$
$=(x-2)(3 x-1)(x+3)$

View full question & answer→

Question 55 Marks

The expression $2x^3 + ax^2 + bx – 2$ leaves remainder $7$ and $0$ when divided by $2x – 3$ and $x + 2$ respectively. Calculate the values of $a$ and $b$

Answer

Left $f(x)=2 x^3+a x^2+b x-2$
$2 x-3=0 \Rightarrow x=\frac{3}{2}$
on dividing f(x) by $2x-3,$ it leaves a remainder $7$
$\therefore 2\left(\frac{3}{2}\right)^3++a\left(\frac{3}{2}\right)^2+b\left(\frac{3}{2}\right)-2=7$
$\frac{27}{4}+\frac{9 a}{4}+\frac{3 b}{2}=9 $
$ \frac{27+9 a+6 b}{4}=9 $
$27+9 a+6 b=36 $
$ 9 a+6 b-9=0 $
$ 3 a+2 b-3=0....(1)$
$x+2=0 \Rightarrow x=-21$
On dividing f(x) by $x+2,$ it leavves a remainder $0.$
$\therefore 2(-2)^3+a(-2)^2+b(-2)-2=0 $
$ -16+4 a-2 b-2=0$
$4 a-2 b-18=0 ......... (2)$
Adding $(1)$ and $(2),$ we get,
$7a-21=0$
$a=3$
Subsituting the value of a in $(1),$ we get,
$3(3)+2b-3 =0$
$9+2b-3=0$
$2b=-6$
$b=-3$

View full question & answer→

Question 65 Marks

If $x^3 + ax^2 + bx + 6$ has $x – 2$ as a factor and leaves a remainder 3 when divided by $x – 3$, find the values of a and b.

Answer

Let $f(x)=x^3+a x^2+b x+6$
$\therefore x-2=0 \Rightarrow x =2$
$(2)^3+a(2)^2+b(2)+6=0$
$8+4 a+2 b+6=0$
$4 a+2 b+14=0$
$2(2 a+b+7)=0$
$2 a+b+7=\frac{0}{2}$
$2 a+b+7=0$
$2 a+b=-7 .... (i)$
$\therefore x-3=0 \Rightarrow x =3$
$(3)^3+a(3)^2+b(3)+6=3$
$27+9 a+3 b+6=3$
$9 a+3 b+33=3$
$9 a+3 b=3-33$
$9 a+3 b=-30$
$3(3 a+b)=-30$
$3 a+b=\frac{-30}{3}$
$3 a+b=-10....(ii)$
Subtracting (i) from (ii), we get,
$2 a+b=-7$
$3 a+b=-10$
$\frac{--\quad+}{-a=3}$
$\therefore a=-3$
Substituting the value of $a = -3$ in (i), we get,
$2a + b = - 7$
$2(-3) + b = - 7$
$- 6 + b + 7 = 0$
$b = - 7 + 6$
$\therefore b = - 1$

View full question & answer→

Question 75 Marks

Given that $x – 2$ and $x + 1$ are factors of $f(x) = x^3 + 3x^2 + ax + b;$ calculate the values of $a$ and $b.$ Hence, find all the factors of $f(x).$

Answer

$f(x)=x^3+3 x^2+a x+b$
Since, $(x-2)$ is a factor of $f(x), f(2)=0$
$\Rightarrow(2)^3+3(2)^2+a(-1)+b=0$
$\Rightarrow 8+12+2 a+b=0$
$\Rightarrow 2 a+b+20=0 \ldots .(1)$
Since, $(x+1)$ is a factor of $f(x), f(-1)=0$
$(-1)^3+3(-1)^2+a(-1)+b=0$
$-1+3-a+b=0$
$-a+b+2=0 \ldots . .(2)$
Subtracting $(2)$ from $(1)$ we get,
$3a+18=0$
$\Rightarrow a=-6$
Subtracting the value of a in $(2),$ we get,
$b=a-2=-6-2=-8$
$\therefore f(x)=x^3+3 x^2-6 x-8$
Now, for $x =-1$
$f(x)=f(-1)=(-1)^3+3(-1)^2-6(-1)-8=-1+3+6-8=0$
Hence, $(x+1)$ is a factor of $f (x)$

$\therefore x^3+3 x^2-6 x-8=(x-1)\left(x^2+2 x-8\right)$
$=(x+1)\left(x^2+4 x-2 x-8\right)$
$=(x+1) \times (x+4)-2(x+4)$
$=(x+1)(x+4)(x-2)$

View full question & answer→

Mathematics [5 marks sum]

Test yourself on this topic