4 Marks Questions

Question 14 Marks

Without actual division, prove that $2x^4 - 5x^3 + 2x^2 - x + 2$ is divisible by $x^2 - 3x + 2.$

Answer

Let $f(x) = 2x^4 - 5x^3 + 2x^2 - x + 2$
$g(x) = x^2 - 3x + 2$
$= x^2 - 2x - x + 2$
$= x(x - 2) - 1(x - 2)$
$= (x - 2)(x - 1)$
Clearly, $(x - 2)$ and $(x - 1)$ are factors of $g(x).$
In order to prove that $f(x)$ is exactly divisible by $g(x),$ it is sufficient to prove that$ f(x)$ is exactly divisible by $(x - 2)$ and $(x - 1).$
Thus, we will show that $(x - 2)$ and $(x - 1)$ are factors of $f(x).$
Now,
$f(2) = 2(2)^4 - 5(2)^3 + 2(2)^2 - 2 + 2$
$= 32 - 40 + 8 = 0$ and
$f(1) = 2(1)^4 - 5(1)^3 + 2(1)^2 - 1 + 2$
$= 2 - 5 + 2 - 1 + 2 = 0$
Therefore, $(x - 2)$ and $(x - 1)$ are factors of $f(x).$
$\Rightarrow g(x) = (x - 2)(x - 1)$ is a factor of $f(x).$
Hence, $f(x)$ is exactly divisible by $g(x).$

View full question & answer→

Question 24 Marks

Using factor theorem, show that $g(x)$ is a factor of $p(x),$ when
$p(x) = 2x^4 + x^3 - 8x^2 - x + 6, g(x) = 2x - 3$

Answer

$f(x) = (2x^4 + x^3 - 8x^2 - x + 6)$
By the Factor Theorem, $(x - a)$ will be a factor of $f(x)$ if $f(a) = 0.$
Here, $2x - 3 = 0$
$\text{x}=\frac{3}{2}$
$\text{f}\Big(\frac{3}{2}\Big)=2\Big(\frac{3}{2}\Big)^4+\Big(\frac{3}{2}\Big)^3-8\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)+6$
$=2\times\frac{81}{16}+\frac{27}{8}-8\times\frac{9}{4}-\frac{3}{2}+6$
$=\frac{81}{8}+\frac{27}{8}-18-\frac{3}{2}+6$
$=\frac{81+27-144-12+48}{8}$
$=\frac{156-156}{8}=0$
$\therefore(2\text{x}-3)$ is a factor of $\big(2\text{x}^4+\text{x}^3-8\text{x}^2-\text{x}+6\big).$

View full question & answer→

Question 34 Marks

Find the values of a and b so that the polynomial $(x^3 - 10x^2 + ax + b)$ is exactly divisible by $(x - 1)$ as well as $(x - 2).$

Answer

Let $f(x) = (x^3 - 10x^2 + ax + b),$ then by factor theorem
$(x - 1)$ and $(x - 2)$ will be factors of $f(x)$ if $f(1) = 0$ and $f(2) = 0.$
$\therefore f(1) = 1^3 - 10 \times 1^2 + a \times 1 + b = 0$
$\Rightarrow 1 - 10 + a + b = 0$
$\Rightarrow a + b = 9 ...(i)$
And $f(2) = 2^3 - 10 \times 2^2 + a \times 2 + b = 0$
$\Rightarrow 8 - 40 + 2a + b = 0$
$\Rightarrow 2a + b = 32 ...(ii)$
Subtracting $(i)$ from $(ii),$ we get
$a = 23$
Substituting the value of $a = 23$ in $(i),$ we get
$23 + b = 9$
$\Rightarrow b = 9 - 23$
$\Rightarrow b = -14$
$\therefore a = 23$ and $b = -14$

View full question & answer→

Question 44 Marks

Without actual division, show that $(x^3 - 3x^2 - 13x + 15)$ is exactly divisible by $(x^2 + 2x - 3).$

Answer

Let $f(x) = x^3 - 3x^2 - 13x + 15$
Now, $x^2 + 2x - 3 = x^2 + 3x - x - 3$
$= x (x + 3) - 1 (x + 3)$
$= (x + 3) (x - 1)$
Thus, $f(x)$ will be exactly divisible by $x^2 + 2x - 3 = (x + 3) (x - 1)$ if $(x + 3)$ and $(x - 1)$ are both factors of $f(x),$ so by factor theorem, we should have $f(-3) = 0$ and $f(1) = 0.$
Now, $f(-3) = (-3)^3 - 3(-3)^2 - 13(-3) + 15$
$= -27 - 3 \times 9 + 39 + 15$
$= -27 - 27 + 39 + 15$
$= -54 + 54 = 0$
And,$ f(1) = 1^3 - 3 \times 1^2 - 13 \times 1 + 15$
$= 1 - 3 - 13 + 15$
$= 16 - 16 = 0$
$\therefore f(-3) = 0$ and $f(1) = 0$
So,$ x^2 + 2x - 3$ divides $f(x)$ exactly.

View full question & answer→

Question 54 Marks

By actual division, find the quotient and the remainder when $(x^4 + 1)$ is divided by $(x - 1).$
Verify that remainder $= f(1).$

Answer

Let $f(x)= x^4 + 1$ and $g(x) = x - 1.$

Quotient $= x^3 + x^2 + x + 1$
Remainder $= 2$
Verification:
Putting $x = 1$ in $f(x),$ we get
$f(1) = 1^4 + 1 = 1 +1 = 2 =$ Remainder, when $f(x) = x^4 + 1$ is divided by $g(x) = x - 1.$

View full question & answer→

MATHS 4 Marks Questions

Test yourself on this topic