Question 14 Marks
In the following, using the remainder theorem, find the remainder when $f(x)$ is divided by $g(x)$ and verify the by actual division:
$f(x) = 9x^3 - 3x^2 + x - 5$, $\text{g(x)}=\text{x}-\frac{2}{3}$
AnswerHere,$f(x) = 9x^3 - 3x^2 + x - 5,$
$\text{g(x)}=\text{x}-\frac{2}{3}$
From, the remainder theorem when $f(x)$ is divided by $\text{g(x)}=\text{x}-\frac{2}{3}$ the remainder will be equal to $\text{f}\Big(\frac{2}{3}\Big)$
Let, $g(x) = 0$
$\Rightarrow\ \text{x}-\frac{2}{3}=0$
$\Rightarrow\ \text{x}=\frac{2}{3}$
Substitute the value of $x$ in $f(x)$
$\text{f}\Big(\frac{2}{3}\Big)=9\Big(\frac{2}{3}\Big)-3\Big(\frac{2}{3}\Big)^2+\Big(\frac{2}{3}\Big)-5$
$=9\Big(\frac{8}{27}\Big)-3\Big(\frac{4}{9}\Big)+\frac{2}{3}-5$
$=\Big(\frac{8}{3}\Big)-\Big(\frac{4}{3}\Big)+\frac{2}{3}-5$
$=\frac{8-4+2-15}{3}$
$=\frac{10-19}{3}$
$=\frac{-9}{3}$
$=-3$
Therefore, the remainder is $-3.$
View full question & answer→Question 24 Marks
Find the value of a, if $x + 2$ is a factor of $4x^4 + 2x^3 - 3x^2 + 8x + 5a$.
AnswerLet $g(x) = x + 2$,
$f(x) = 4x^4 + 2x^3 - 3x^2 + 8x + 5a.$
Let $g(x) = 0$
$\Rightarrow x + 2 = 0 $
$\Rightarrow x = -2,$
$\because$ $g(x)$ is a factor of $f(x)$
$\therefore$ $f(-2) = 0$
$f(-2) = 4(-2)^4 + 2(-2)^3 - 3(-2)^2 + 8(-2) + 5a = 0 $
$\Rightarrow 4(16) + 2(-8) - 3(4) + 8(-2) + 5a = 0 $
$\Rightarrow 64 - 16 - 12 - 16 + 5a = 0 $
$\Rightarrow 20 + 5a = 0 $
$\Rightarrow 5a = -20$
$\Rightarrow\ \text{a}=\frac{-20}{5}=-4$
$\therefore\ \text{a}=-4$
View full question & answer→Question 34 Marks
Using factor theorem, factorize the following polynomials:
$2y^3 + y^2 - 2y - 1$
AnswerLet $p(y) = 2y^3 + y^2 - 2y - 1$ By hit and trial method
$p(1) = 2(1)^3 + (1)^2 - 2(1) - 1 = 2 + 1 - 2 - 1 = 0$
So, $y -1$ is a factor of this polynomial.
Now, By long division method,
$\therefore$ $2y^3 + y^2 - 2y - 1 = (y - 1)(2y^2 + 3y + 1)$
$= (y - 1)(2y^2 + 2y + y + 1)$
$= (y - 1)[2y(y + 1) + 1(y + 1)]$
$= (y - 1)(y + 1)(2y + 1)$ View full question & answer→Question 44 Marks
If $\text{x}=-\frac{1}{2}$ is zero of the polynomial $p(x) = 8x^3 - ax^2 - x + 2$, Find the value of a.
AnswerWe know that, $p(x) = 8x^3 - ax^2 - x + 2$
Given that the value of $\text{x}=-\frac{1}{2}$
Substitute the value of x in f(x)$\text{p}\Big(-\frac{1}{2}\Big)=8\Big(-\frac{1}{2}\Big)^3-\text{a}\Big(-\frac{1}{2}\Big)^2-\Big(-\frac{1}{2}\Big)+2$
$=-8\Big(\frac{1}{8}\Big)-\text{a}\Big(\frac{1}{4}\Big)+\Big(\frac{1}{2}\Big)+2$
$=-1-\Big(\frac{\text{a}}{4}+\frac{1}{2}+2\Big)$
$=1-\Big(\frac{\text{a}}{4}+\frac{1}{2}\Big)$
$=\frac{3}{2}-\frac{\text{a}}{4}$
To, find the value of a, equal $\text{p}\Big(-\frac{1}{2}\Big)$ to zero$\text{p}\Big(-\frac{1}{2}\Big)=0$
$\frac{3}{2}-\frac{\text{a}}{4}=0$
On taking L.C.M$\frac{6-\text{a}}{4}=0$
$\Rightarrow6-\text{a}=0$
$\Rightarrow\text{a}=6$
View full question & answer→Question 54 Marks
In the following, using the remainder theorem, find the remainder when $f(x)$ is divided by $g(x)$ and verify the by actual division:
$f(x) = x^3 + 4x^2 - 3x + 10, g(x) = x + 4$
AnswerHere,
$f(x) = x^3 + 4x^2 - 3x + 10$
$g(x) = x + 4$
From, the remainder theorem when $f(x$) is divided by $g(x) = x - (-4)$ the remainder will be equal to $f(-4)$
Let, $g(x) = 0$
$\Rightarrow x + 4 = 0$
$\Rightarrow x = -4$
Substitute the value of $x$ in $f(x)$
$f(-4) = (-4)^3 + 4(-4)^2- 3(-4) + 10$
$= - 64 + (4 \times 16) + 12 + 10$
$= - 64 + 64 + 12 + 10$
$= 12 + 10$
$= 22$
Therefore, the remainder is $22.$
View full question & answer→Question 64 Marks
In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:$\text{f(x)}=3\text{x}^4+2\text{x}^3-\frac{\text{x}^3}{3}-\frac{\text{x}}{9}+\frac{2}{27},$ $\text{g(x)}=\text{x}+\frac{2}{3}$
AnswerHere,$\text{f(x)}=3\text{x}^4+2\text{x}^3-\frac{\text{x}^3}{3}-\frac{\text{x}}{9}+\frac{2}{27},$
$\text{g(x)}=\text{x}+\frac{2}{3}$
From, the remainder theorem when f(x) is divided by $\text{g(x)}=\text{x}-\Big(-\frac{2}{3}\Big),$ the remainder will be equal to $\text{f}\Big(-\frac{2}{3}\Big)$ Substitute the value of x in f(x)$\text{f}\Big(-\frac{2}{3}\Big)=3\Big(-\frac{2}{3}\Big)^4+2\Big(-\frac{2}{3}\Big)^3\\-\frac{\Big(-\frac{2}{3}\Big)^3}{3}-\Bigg[\frac{\big(-\frac{2}{3}\big)}{9}+\frac{22}{7}\Big(\frac{2}{27}\Big)\Bigg]$
$=3\Big(\frac{16}{81}\Big)+2\Big(\frac{-8}{27}\Big)-\frac{4}{(9\times3)}-\Big(\frac{-2}{(9\times3)}\Big)+\frac{2}{27}$
$=\Big(\frac{16}{27}\Big)-\Big(\frac{16}{27}\Big)-\frac{4}{27}+\Big(\frac{2}{27}\Big)+\frac{2}{27}$
$=\Big(\frac{4}{27}\Big)-\Big(\frac{4}{27}\Big)$
$=0$
Therefore, the remainder is 0.
View full question & answer→Question 74 Marks
Using factor theorem, factorize the following polynomials:
$3x^3 - x^2 - 3x + 1$
AnswerLet $f(x) = 3x^3 - x^2 - 3x + 1$
The factor of the coefficient of $x^3$ is $3$.
So, the possible rational roots of $f(x)$ are $\pm1$ and $\pm\frac{1}{3}$
We have,
$f(1) = 3 - 1 - 3 + 1 = 0$
$\Rightarrow (x - 1)$ is a factor of $f(x)$
$f(-1) = -3 - 1 + 3 + 1$
$\Rightarrow (x + 1)$ is a factor of $f(x)$
So, $(x - 1)$ and $(x + 1)$ are factors of $f(x)$
$\Rightarrow (x - 1)(x + 1)$ is a also a factor of $f(x)$
$\Rightarrow x^2 - 1$ is a factor of $f(x)$.
Let us now divide $f(x) = 3x^3 - x^2 - 3x + 1$ by $x^2 - 1$ to get the other factors of $f(x).$
By long division, we have:
Therefore,
$3x^3 - x^2 - 3x + 1 = (x^2 - 1)(3x - 1)$
Now,
$(x^2 - 1) = (x - 1)(x + 1)$
Hence,
$3x^3 - x^2 - 3x + 1 = (x - 1)(x + 1)(3x - 1)$ View full question & answer→Question 84 Marks
What must be added to $x^3 - 3x^2 - 12x + 19$ so that the result is exactly divisibly by $x^2 + x - 6$?
AnswerLet $p ( x )= x ^3-3 x ^2-12 x +19$ and $q ( x )= x ^2+ x -6$
When $p(x)$ is divided $q(x)$ then the remainder is a linear equation, $r(x)$.
Let $r ( x )= ax + b$ when added to $p ( x )$ we get an expression which is divisible by $q ( x )$.
$f(x)=p(x)+r(x) f(x)$
$=x^3-3 x^2-12 x+19+a x+b$
$=x^3-3 x^2-12 x+a x+19+b$
$=x^3-3 x^2-x(12-a)+19+b$
Now, $q(x)=x^2+x-6=x^2+3 x-2 x-6=(x+3)(x-2)$
Also, $f(x)$ is divisible $q(x) \therefore f(-3)=0, f(2)=0$
$f(-3)=(-3)^3-3(-3)^2-12(-3)+19+a(-3)+b=0$
$\Rightarrow-27+27+36+19-3 a+b=0$
$\Rightarrow b=3 a+27+27-36-19$
$\Rightarrow b=3 a-1 \ldots(1)$
$f(2)=2^3-3(2)^2-12(2)+19+2 a+b=0$
$\Rightarrow 8-12-24+19+2 a+b=0$
$\Rightarrow b=9-2 a \ldots(2) \text { Equating (1) and (2) } 3 a-1=9-2 a$
$\Rightarrow 5 a=10$
$\Rightarrow a=2$
$\therefore b=3 a-13(2)-1=5$ [Substituting $b=5$, in equation $1$]
Hence, $a x+b=2 x+5$
View full question & answer→Question 94 Marks
Using factor theorem, factorize the following polynomials:
$x^4 - 7x^3_+ 9x^2 + 7x - 10$
AnswerLet $f(x)=x^4-7 x^3+9 x^2+7 x-10$
The factors of constant term in $f(x)$ are $\pm 1, \pm 2, \pm 5$ and $\pm 10$.
We have,
$f(1)=1-7+9+7-10=0$
$\Rightarrow( x -1)$ is a factor of $f ( x )$
$f(-1)=1+7+9-7-10=0$
$\Rightarrow(x+1)$ is a factor of $f(x)$
$f(2)=16-56+36+14-10=0$
$\Rightarrow( x -2)$ is a factor of $f ( x )$
$f(-2)=16+56-36-14-10=10$
$\Rightarrow(x+2)$ is not a factor of $f(x)$
$f(5)=625-875+225+35-10=0$
$\Rightarrow(x-5)$ is a factor of $f(x)$
Since $f(x)$ is a polynomial of degree 4. So, it cannot have more than 4 linear factors.
Thus, factors of $f(x)$ are $(x-1),(x+1),(x-2)$ and $(x-5)$.
Therefore,
$f(x)=k(x-1)(x+1)(x-2)(x-5)$
$x^4-7 x^3+9 x^2+7 x-10=k(x-1)(x+1)(x-2)(x-5)$
Putting $x=0$ on both sides, we get,
$-10=k(-1)(1)(-2)(-5)$
$-10=-10 k$
$k=1$
Substituting $k =1$ in $(1)$, we get,
$x^4-7 x^3+9 x^2+7 x-10=(x-1)(x+1)(x-2)(x-5)$
View full question & answer→Question 104 Marks
Using factor theorem, factorize the following polynomials:
$x^3 - 23x^2 + 142x - 120$
AnswerLet $p(x)=x^3-23 x^2+142 x-120$
We shall now look for all the factors of $-120 .$
Some of these are $1, \pm 2, \pm 3, \pm, \pm 4, \pm 5, \pm 6, \pm 8$
$\pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm, 60$
By trial, we find that $p(1)=0$.
So $x-1$ is a factor of $p(x)$.
Now, we see that $x^3-23 x^2+142 x-120=x^3-x^2-22 x^2+22 x+120 x-120$
$=x^2(x-1)-22 x(x-1)+120(x-1)$
$=(x-1)\left(x^2-22 x+120\right)[\text { Taking }(x-1) \text { common }]$
We could have also got this by dividing $p(x)$ by $x-1$.
Now $x^2-22 x+120$ can be factorised either by splitting the middle term or by using the factor theorem.
By splitting the middle term, we have:
$x^2-22 x+120=x^2-12 x-10 x+120$
$=x(x-12)-10(x-12)$
$=(x-12)(x-10)$
$\text { So, } x^3-23 x^2+142 x-120$
$=(x-1)(x-10)(x-12)$
View full question & answer→Question 114 Marks
In the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not:
$f(x) = x^3 - 6x^2 + 11x - 6, $
$g(x)=x^2 - 3x + 2$
Answer$g(x)=x^2-3 x+2$
$=x^2-2 x-x+2$
$=x(x-2)-1(x-2)$
$=(x-2)(x-1)$
Let $g(x)=0$
$\Rightarrow(x-2)(x-1)=0 \text { If } x-2=0$
$\Rightarrow x=2 \text { If } x-1=0$
$\Rightarrow x=1$
$f(2)=2^3-6(2)^2+11(2)-6$
$=8-24+22-6=30-30$
$=0 f(1)=1^3-6(1)^2+11(1)-6$
$=1-6+11-6=0$
$\because f(2)=0$ and $f(1)=0$, by factor theorem, $(x-2)$ and $(x-1)$ both are factors of $f(x)$.
Hence, $(x-2)(x-1)$ is a factor of $f(x)$.
View full question & answer→Question 124 Marks
Using factor theorem, factorize the following polynomials:
$y^3 - 7y + 6$
AnswerLet $f(y)=y^3-7 y+6$
The factors of constant term in $f(y)$ are $\pm 1, \pm, 2, \pm 3$ and $\pm 6$.
We have,
$f (1)=1-7+6=0$
$\Rightarrow( y -1)$ is a factor of $f ( y )$
$f (-1)=-1+7+6=12$
$\Rightarrow(y+1)$ is a factor of $f(y)$
$f(2)=8-14+6=0$
$\Rightarrow( y -2)$ is a factor of $f ( y )$
$f(-2)=-8+14+6=12$
$\Rightarrow(y+2)$ is not a factor of $f ( y )$
$f(3)=27-21+6=12$
$\Rightarrow(y-3)$ is not a factor of $f(y)$
$f(-3)=-27+21+6=0$
$\Rightarrow(y+3)$ is a factor of $f(y)$
Since $f(y)$ is a polynomial of degree 3 . So, it cannot have more than 3 linear factors.
Thus, factors of $f(y)$ are $(y-1)(y-2)$ and $(y+3)$.
Therefore,
$f(y)=k(y-1)(y-2)(y+3)$
$y^3-7 y+6=k(y-1)(y-2)(y+3) \ldots(1)$
Putting $y=0$ on both sides, we get,
$6=k(-1)(-2)(3)$
$6=6 k$
$k=1$
Substituting $k=1$ in (1), we get,
$y^3-7 y+6=(y-1)(y-2)(y+3)$
View full question & answer→Question 134 Marks
Find the values of a and b, if $x^2 - 4$ is a factor of $ax^4 + 2x^3 - 3x^2 + bx - 4.$
AnswerLet $g(x)=x^2-4$ is, $f(x)=a x^4+2 x^3-3 x^2+b x-4$.
Let $g(x)=0$
$\Rightarrow x ^2-4=0$
$\Rightarrow x^2=4$,
$\Rightarrow x = \pm 2$
Since $\left(x^2-4\right)$ is a factor of $f(x)$.
$\therefore f(2)=0 \text { and } f(-2)=0$
$f(2)=a(2)^4+2(2)^3-3(2)^2+b(2)-4=0$
$\Rightarrow 16 a+16-12+2 b-4=0$
$\Rightarrow 16 a+2 b=0$
$\Rightarrow 16 a=-2 b$
$\Rightarrow a=\frac{-2 b}{16}=\frac{-b}{8} \ldots(1)$
Also, $f(-2)=0 f(-2)=a(-2)^4+2(-2)^3-3(-2)^2+b(-2)-4=0$
$\Rightarrow 16 a-16-12-2 b-4=0$
$\Rightarrow 16 a-2 b-32=0$
$\Rightarrow 16 a=2 b+32$
$\Rightarrow \frac{2 b+32}{16} \ldots(2)$
Equating equations (1) and (2) $\Rightarrow \frac{- b }{8}=\frac{2 b+32}{16}$
$\Rightarrow \frac{-16 b}{8}=2 b+32$
$\Rightarrow-2 b=2 b+32$
$\Rightarrow-4 b=32$
$\Rightarrow b=-8$
Substituting $b =-8$ in equation (1) $a =\frac{- b }{8}=\frac{-(-8)}{8}=\frac{8}{8}=1$
$\therefore a=1, b=-8$
View full question & answer→Question 144 Marks
In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
$f(x) = 2x^4 - 6x^3 + 2x^2 - x + 2, g(x) = x + 2$
AnswerHere,
f(x) $= 2x^4 - 6x^3 + 2x^2 - x + 2$
g(x) $= x + 2$
From, the remainder theorem when $f(x)$ is divided by $g(x) = x - (-2)$ the remainder will be equal to $f(-2)$
Let, $g(x) = 0$
$\Rightarrow x + 2 = 0$
$\Rightarrow x = -2$
Substitute the value of x in f(x)
$f(-2) = 2(-2)^4 - 6(-2)^3 + 2(-2)^2 - (-2) + 2$
$= (2 \times 16) - (6 \times (-8)) + (2 \times 4) + 2 + 2$
$= 32 + 48 + 8 + 2 + 2$
$= 92$
Therefore, the remainder is 92.
View full question & answer→Question 154 Marks
$2x^4 - 7x^3 - 13x^2 + 63x - 45$
AnswerLet $f(x)=2 x^4-7 x^3-13 x^2+63 x-45$ be the given polynomial.
Now, putting $x=1$, we get
$f(1)=2(1)^4-7(1)^3-13(1)^2+63(1)-45$
$=2-7-13+63-45=65-65$
$=0$
Therefore, $(x-1)$ is a factor of polynomial $f(x)$.
Now,
$f(x)=2 x^3(x-1)-5 x^2(x-1)-18 x(x-1)+45(x-1)$
$=(x-1)\left(2 x^3-5 x^2-18 x+45\right)$
$=(x-1) g(x) \ldots(1)$
Where $g(x)=2 x^3-5 x^2-18 x+45$
Putting $x=3$, we get:
$g(3)=2(3)^3-5(3)^2-18(3)+45$
$=54-45-54+45$
$=0$
Therefore, $( x -3)$ is the factor of $g ( x )$.
Now,
$g(x)=2 x^2(x-3)+x(x-3)-15(x-3)$
$=(x-3)\left(2 x^2+x-15\right)$
$=(x-3)\left(2 x^2+6 x-5 x-15\right)$
$=(x-3)(x+3)(2 x-5) \ldots(2)$
From equation (1) and (2), we get:
$f(x)=(x-1)(x-3)(x+3)(2 x-5)$
Hence, $(x-1),(x-3),(x+3)$ and $(2 x-5)$ are the factors of polynomial $f(x)$.
View full question & answer→Question 164 Marks
Using factor theorem, factorize the following polynomials:
$x^3 + 6x^2 + 11x + 6$
AnswerLet $x = 1$
$\text{f(1)}=1^3+6(2)^2+11(1)+6\neq0$
Let $x = -1 f(-1)$
$= (-1)^3 + 6(-1)^2 + 11(-1) + 6 = 12 - 12 = 0$
$\therefore$ $x = -1$ is a solution
$\Rightarrow x + 1 = 0$ i. e $(x + 1)$ is a factor of $f(x)$
By division algorithm $x^3 + 6x^2 + 11x + 6$
$= (x + 1)(x^2 + 5x + 6)$
$= (x + 1)(x^2 + 2x + 3x + 6)$
$= (x + 1)(x(x + 2) + 3(x + 2))$
$= (x + 1)(x + 2)(x + 3)$
$\therefore$ $x^3 + 6x^2 + 11x + 6$
$= (x + 1)(x + 2)(x + 3)$ View full question & answer→Question 174 Marks
$x^4 - 2x^3 - 7x^2 + 8x + 12.$
AnswerLet $f(x)=x^4-2 x^3-7 x^2+8 x+12$
The factors of constant term in $f(x)$ are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6$ and $\pm 12$.
We have,
$f(1)=1-2-7+8+12=12$
$\Rightarrow( x -1)$ is not a factor of $f ( x )$
$f(-1)=1+2-7-8+12=0$
$\Rightarrow(x+1)$ is a factor of $f(x)$
$f(2)=16-16-28-16+12=0$
$\Rightarrow( x -2)$ is a factor of $f ( x )$
$f(-2)=16+16-28-16+12=0$
$\Rightarrow(x+2)$ is a factor of $f(x)$
$f(3)=81-54-63+24+12=0$
$\Rightarrow(x-3)$ is a factor of $f(x)$
Since $f(x)$ is a polynomial of degree 4 . So, it cannot have more than 4 linear factors.
Thus, factors of $f(x)$ are $(x+1),(x-2),(x+2)$ and $(x-3)$.
Therefore,
$f(x)=k(x+1)(x+2)(x-2)(x-3)$
$x^4-2 x^3-7 x^2+8 x+12=k(x+1)(x+2)(x-2)(x-3)$
Putting $x=0$ on both sides, we get,
$12=k(1)(2)(-2)(-3)$
$12=12 k$
$k=1$
Substituting $k =1$ in (1), we get,
$x^4-2 x^3-7 x^2+8 x+12=k(x+1)(x+2)(x-2)(x-3)$
View full question & answer→Question 184 Marks
Using factor theorem, factorize the following polynomials:
$x^3 - 3x^2 - 9x - 5$
AnswerLet $p(x) = x^3 - 3x^2 - 9x - 5$
The factors of 5 are $\pm1,\pm5.$
By hit and trial method $p(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3 + 9 - 5 = 0$
So $x + 1$ is a factor of this polynomial p(x).
Let us find the quotient while dividing $x^3 - 3x^2 - 9x - 5$ by $x + 1$ By long division
Now, Dividend = $Divisor \times Quotient + Remainder$
$\therefore$$ x^3 - 3x^2 - 9x - 5$
$= (x + 1)(x^2 - 4x - 5) + 0$
$= (x + 1)(x^2 - 5x + x - 5)$
$= (x + 1)[x(x - 5) + 1(x - 5)]$
$= (x + 1)(x - 5)(x + 1) = (x - 5)(x + 1)(x + 1)$ View full question & answer→Question 194 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$p(x) = x^3 - 6x^2 + 11x - 6, x = 1, 2, 3$
Answerp(x) = $x^3 - 6x^2 + 11x - 6, x = 1, 2, 3$
We know that,
p(x) = $x^3 - 6x^2 + 11x - 6$
given that the values of x are $1, 2 , 3$
substitute $x = 1$ in $p(x)$
$p(1) = 1^3- 6(1)^2 + 11(1) - 6$
$= 1 - (6 \times 1) + 11 - 6$
$= 1 - 6 + 11 - 6$
$= 0$
Now, substitute $x = 2$ in $p(x)$
$P(2) = 2^3 - 6(2)^2 + 11(2) - 6$
$= (2 \times 3) - (6 \times 4) + (11 \times 2) - 6$
$= 8 - 24 - 22 - 6$
$= 0$
Now, substitute $x = 3$ in $p(x)$
$P(3) = 3^3- 6(3)^2 + 11(3) - 6$
$= (3 \times 3) - (6 \times 9) + (11 \times 3) - 6$
$= 27 - 54 + 33 - 6$
$= 0$
Since, the result is 0 for $x = 1, 2, 3$ these are the roots of $x^3 - 6x^2 + 11x - 6$
View full question & answer→Question 204 Marks
If the polynomials $a x^3+3 x^2-13$ and $2 x^3-5 x+a$, when divided by $(x-2)$ leave the same remainder, Find the value of $a$.
AnswerHere,
The polynomials are:
$f(x)=a x^3+3 x^2-13$
$p(x)=2 x^3-5 x+a$
$\text { equate, } x-2=0$
$x=2$
substitute the value of $x$ in $f(x)$ and $p(x)$
$f(2)=(2)^3+3(2)^2-13$
$=8 a+12-13$
$=8 a-1 \ldots(1)$
$p(2)=2(2)^3-5(2)+a$
$=16-10+a$
$=6+a \ldots(2)$
$f(2)=p(2)$
$\Rightarrow 8 a-1=6+a$
$\Rightarrow 8 a-a=6+1$
$\Rightarrow 7 a=7$
$\Rightarrow a=1$
The value of $a=1$.
View full question & answer→Question 214 Marks
Find the rational roots of the polynomial $f(x) = 2x^3 + x^2 - 7x - 6$
AnswerGiven that $f(x)=2 x^3+x^2-7 x-6 f(x)$ is a cubic polynomial with an integer coefficient.
If the rational root in the form of $\frac{p}{q}$,
the values of $p$ are limited to factors of 6 which are $\pm 1, \pm 2, \pm 3, \pm 6$ and the values of $q$ are limited to the highest degree coefficient i.e 2 which are $\pm 1, \pm 2$ here,
the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$
$\text { Let, } x=-1 f(-1)$
$=2(-1)^3+(-1)^2-7(-1)-6$
$=-2+1+7-6$
$=-8+8=0$
Let, $x=2 f (-2)$
$=2(2)^3+(2)^2-7(2)-6$
$=(2 \times 8)+4-14-6$
$=16+4-14-6$
$=20-20=0$
Let, $x =-\frac{3}{2}$
$f =\left(-\frac{3}{2}\right)=2\left(-\frac{3}{2}\right)^3+\left(-\frac{3}{2}\right)^2-7\left(-\frac{3}{2}\right)-6$
$=2\left(-\frac{27}{4}\right)+\frac{9}{4}-7\left(-\frac{3}{2}\right)-6$
$=2\left(-\frac{27}{4}\right)+\frac{9}{4}-\left(-\frac{21}{2}\right)-6$
$=-6.75+2.25+10.5-6$
$=12.75-12.75$
$=0$
But from all the factors only $-1,2$ and $-\frac{3}{2}$ gives the result as zero
So, the rational roots of $2 x^3+x^2-7 x-6$ are $-1,2$ and $-\frac{3}{2}$
View full question & answer→Question 224 Marks
Using factor theorem, factorize the following polynomials:
$x^3 + 2x^2 - x - 2$
AnswerLet $x=1 f(1)=1^3+2(1)^2-1-2=0$
$\therefore x=1$ is a solution
$\Rightarrow x -1=0$ i. e $( x -1)$ is a factor of $f ( x )$
By division algorithm $x^3+2 x^2-x-2$
$=(x-1)\left(x^2+3 x+2\right)$
$=(x-1)\left(x^2+2 x+x+2\right)$
$=(x-1)(x(x+2)+1(x+2))$
$=(x-1)(x+2)(x+1)$
$\therefore x^3+2 x^2-x-2=(x-1)(x+2)(x+1)$ View full question & answer→Question 234 Marks
Factorize the following polynomials:
$x^3+13 x^2+31 x-45$ given that $x+9$ is a factor.
AnswerLet $f(x)=x^3+13 x^2+31 x-45$ be the given polynomial.
Therefore, $(x+9)$ is a factor of the polynomial $f(x)$.
Now,
$f(x)=x^2(x+9)+4 x(x+9)-5(x+9)$
$=(x+9)\left(x^2+4 x-5\right)$
$=(x+9)\left(x^2+5 x-x-5\right)$
$=(x+9)(x-1)(x+5)$
Hence $(x-1),(x+5)$ and $(x+9)$ are the factors of polynomial $f(x)$.
View full question & answer→Question 244 Marks
If both $x+1$ and $x-1$ are factors of $a x^3+x^2-2 x+b$, find the values of $a$ and $b$.
AnswerLet, $x+1=0 x=-1 \because(x+1)$ is a factor of $p(x)=a x^3+x^2-2 x+b$
$\therefore p(-1)=0$
$p(-1)=a(-1)^3+(-1)^2-2(-1)+b$
$=0 \Rightarrow-a+1+2+b$
$=0 \Rightarrow-a+3+b$
$=0 \Rightarrow a=3+b$
Let, $x -1=0 x =1$
$\because(x-1)$ is a factor of $p(x)$
$\therefore p(1)=0$
$p(1)=a(1)^3+1^2-2(1)+b$
$=0 \Rightarrow a+1-2+b$
$=0 \Rightarrow a=-b+1 \ldots(2) \text { Equating (1) and (2) }$
$\Rightarrow 3+b=-b+1 \Rightarrow b+b=1-3 \Rightarrow 2 b=-2 \Rightarrow b=-1$
Substituting $b=-1$ in equation (2) $a=-(-1)+1=1+1=2$
$\therefore a=2, b=-1$
View full question & answer→Question 254 Marks
Using factor theorem, factorize the following polynomials:
$x^3 - 2x^2 - x + 2$
AnswerLet $f(x)=x^3-2 x^2-x+2$
The factors of constant term in $f(x)$ are $\pm 1, \pm 2$.
We have
$f(1)=1-2-1+2=0$
$\Rightarrow( x -1)$ is a factor of $f ( x )$
$f(-1)=-1-2+1+2=0$
$\Rightarrow( x +1)$ is a factor of $f ( x )$
$f(2)=8-8-2+2=0$
$\Rightarrow( x -2)$ is a factor of $f ( x )$
Since $f(x)$ is a polynomial of degree 3 . So, it cannot have more than 3 linear factors.
Thus, factors of $f(x)$ are $(x-1),(x+1)$ and $(x-2)$.
Therefore,
$f(x)=k(x-1)(x+1)(x-2)$
$x^3-2 x^2-x+2=k(x-1)(x+1)(x-2) \ldots$
Putting $x=0$ on both sides, we get,
$2=k(-1)(1)(-2)$
$2=2 k$
$k=1$
Substituting $k =1$ in (1), we get,
$x^3-2 x^2-x+2=(x-1)(x+1)(x-2)$
View full question & answer→Question 264 Marks
If the polynomial $2x^3 + ax^2 + 3x - 5$ and $x^3 + x^2 - 4x + a$ leave the same remainder when divided by $x - 2$, Find the value of a.
AnswerGiven, the polymials are:
$f(x)=2 x^3+a x^2+3 x-5$
$p(x)=x^3+x^2-4 x+a$
The remainders are $f(2)$ and $p(2)$ when $f(x)$ and $p(x)$ are divided by $x-2$
We know that, $f(2)=p(2)$ (given in problem)
we need to calculate $f(2)$ and $p(2)$ for, $f(2)$
substitute $(x=2)$ in $f(x) f(2)$
$=2(2)^3+a(2)^2+3(2)-5$
$=(2 \times 8)+4 a+6-5=16+4 a+1=4 a+17 \ldots(1) \text { for, } p(2)$
$\text { substitute }(x=2) \text { in } p(x) p(2)=2^3+2^2-4(2)+a$
$=8+4-8+a=4+a \ldots(2) \text { Since, } f(2)=p(2)$
Equate equation 1 and 2
$\Rightarrow 4 a+17=4+a$
$\Rightarrow 4 a-a=4-17$
$\Rightarrow 3 a=-13$
$\Rightarrow a=\frac{-13}{3}$
The value of $a =\frac{-13}{3}$.
View full question & answer→Question 274 Marks
In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
$f(x) = x^4 - 3x^2 + 4, g(x) = x - 2$
AnswerHere,
$f(x)=x^4-3 x^2+4$
$g(x)=x-2$
From, the remainder theorem when $f(x)$ is divided by $g(x)=x-2$ the remainder will be equal to $f(2)$
Let, $g(x)=0$
$\Rightarrow x-2=0$
$\Rightarrow x=2$
Substitute the value of $x$ in $f(x)$
$f(2)=2^4-3(2)^2+4$
$=16-(3 \times 4)+4$
$=16-12+4$
$=20-12$
$=8$
Therefore, the remainder is 8 .
View full question & answer→Question 284 Marks
Using factor theorem, factorize the following polynomials:
$y^3 - 2y^2 - 29y - 42$
AnswerLet $f(y) = y^3 - 2y^2 - 29y - 42$ be the given polynomial.
Now, putting $y = -2$, we get
$f(-2) = (-2)^3 - 2(-2)^2 - 29(-2) - 42$
$= -8 - 8 + 58 - 42$
$= -58 + 58 = 0$
Therefore, $(y + 2)$ is a factor of polynomial f(y).
Now,
$f(y) = y^2(y + 2) + 4y(y + 2) - 2(y + 2)$
$= (y + 2)(y^2 - 4y - 21)$
$= (y + 2)(y^2 - 7y + 3y - 21)$
$= (y + 2)(y + 3)(y - 7)$
Hence$ (y + 2), (y + 3)$ and $(y - 7)$ are the factors of polynomial f(y).
View full question & answer→Question 294 Marks
Show that $(x - 2), (x + 3)$ and $(x - 4)$ are factors of $x^3 - 3x^2 - 10x + 24.$
Answer$F(x)=x^3-3 x^2-10 x+24$
$\text { Let, } x-2=0$
$\Rightarrow x=2 f(2)$
$=2^3-3(2)^2-10(2)+24$
$=8-12-20+24$
$=32-32=0$
$\text { Let, } x+3=0$
$\Rightarrow x=-3 f(-3)$
$=(-3)^3-3(-3)^2-10(-3)+24$
$=-27-3(9)+30+24$
$=-27-27+30+24$
$=-54+54=0 L$
$\text { et, } x-4=0$
$\Rightarrow x=4 f(4)$
$=4^3-3(4)^2-10(4)+24$
$=64-3(16)-40+24$
$=64-48-40+24$
$=88-88=0$
$\because f(2)=0, f(-3)=0, f(4)=0$
$\therefore$ By factore theoram $( x -2),( x +3)$ and $( x -4)$ are factors of $x ^3-3 x ^2-10 x +24$.
View full question & answer→Question 304 Marks
Using factor theorem, factorize the following polynomials:
$x^3 + 13x^2 + 32x + 20$
AnswerLet $p(x) = x^3 + 13x^2+ 32x + 20$
The factors of 20 are $\pm1,\pm2,\pm4,\pm5\dots$
By hit and trial method
$p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20$
$= -1 + 13 - 32 + 20$
$= 33 - 33 = 0$
As $p(-1$) is zero, so $x + 1$ is a factor of this polynomial p(x).
Let us find the quotient while dividing $x^3 + 13x^2 + 32x + 20 by (x + 1)$
By long division
We know that
Dividend = $Divisor \times Quotient + Remainder$
$x^3 + 13x^2 + 32x + 20 = (x + 1)(x^2 + 12x + 20) + 0$
$= (x + 1)(x^2 + 10x + 2x + 20)$
$= (x + 1)[x(x + 10) + 2(x + 10)]$
$= (x + 1)(x + 10)(x + 2)$
$= (x + 1)(x + 2)(x + 10)$ View full question & answer→Question 314 Marks
In the following, using the remainder theorem, find the remainder when $f(x)$ is divided by $g(x)$ and verify the by actual division:
$f(x) = 4x^4 - 3x^3 - 2x^2 + x - 7, g(x) = x - 1$
AnswerHere,
$f(x)=4 x^4-3 x^3-2 x^2+x-7$
$g(x)=x-1$
From, the remainder theorem when $f(x)$ is divided by $g(x)=x-(-1)$ the remainder will be equal to $f(1)$
Let, $g(x)=0$
$\Rightarrow x-1=0$
$\Rightarrow x=1$
Substitute the value of $x$ in $f(x)$
$f(1)=4(1)^4-3(1)^3-2(1)^2+1-7$
$=4-3-2+1-7$
$=5-12$
$=-7$
Therefore, the remainder is $7 .$
View full question & answer→Question 324 Marks
$x^4 + 10x^3 + 35x^2 + 50x + 24.$
AnswerLet $f(x)=x^4+10 x^3+35 x^2+50 x+24$
Now, putting $x=-1$, we get
$f(-1)=(-1)^4+10(-1)^3+35(-1)^2+50(-1)+24$
$=1-10+35-50+24=60-60$
$=0$
Therefore, $(x+1)$ is a factor of polynomial $f(x)$.
Now,
$f(x)=x^3(x+1)+9 x^2(x+1)+26(x+1)+24(x+1)$
$=(x+1)\left(x^3+9 x^2+26 x+24\right)$
$=(x+1) g(x) \ldots(1)$
Where $g(x)=x^3+9 x^2+26 x+24$
Putting $x=-2$, we get:
$g(-2)=(-2)^3+9(-2)^2+26 x(-2)+24$
$=-8+36-52+24=60-60$
$=0$
Therefore, $(x+2)$ is the factor of $g(x)$.
Now,
$g(x)=x^2(x+2)+7 x(x+2)+12(x+2)$
$=(x+2)\left(x^2+7 x+12\right)$
$=(x+2)\left(x^2+4 x+3 x+12\right)$
$=(x+2)(x+3)(x+4) \ldots(2)$
From equation (1) and (2), we get:
$f(x)=(x+1)(x+2)(x+3)(x+4)$
Hence,
$(x+1),(x+2),(x+3)$ and $(x+4)$ are the factors of polynomial $f(x)$.
View full question & answer→Question 334 Marks
If $x^3+a x^2-b x+10$ is divisible by $x^2-3 x+2$, find the values of $a$ and $b$.
AnswerLet, $p(x)=x^3+a x^2-bx+10$
$g(x)=x^2-3 x+2$
$\text { Let } g(x)=0$
$\Rightarrow x^2-3 x+2=0$
$\Rightarrow x^2-2 x-x+2$
$\Rightarrow x(x-2)-1(x-2)=0$
$\Rightarrow x-2=0, x-1=0$
$\therefore x=2, x=1$
Since $x^2-3 x+2$ is a factor of $p(x) \therefore(x-2)(x-1)$ is a factor of $p(x)$
Hence, $p(2)=0, p(1)=0 p(2)=2^3+a(2)^2-b(2)+10=0$
$\Rightarrow 8+4 a-2 b+10=0$
$\Rightarrow 4 a-2 b+18=0$
$\Rightarrow 4 a=-18+2 b$
$\Rightarrow a=\frac{-18+2 b}{4}=\frac{2(-9+b)}{4}=\frac{-9+b}{2} \ldots(1)$
$p(1)=13+a(1) 2-b(1)+10=0$
$\Rightarrow a-b+11=0$
$\Rightarrow a=b-11 \ldots \text { (2) Equadting (1) and (2) } \frac{-9+b}{2}=b-11$
$\Rightarrow-9+b=2 b-22$
$\Rightarrow-9+22=2 b-b$
$\Rightarrow 13=b$
Subsitituting $b =13$ in equation (2) $a = b -11=13-11=2$
$\therefore a=2, b=13$
View full question & answer→Question 344 Marks
Using factor theorem, factorize the following polynomials:
$2y^3 - 5y^2 - 19y + 42$
AnswerLet $f(y)=2 y^3-5 y^2-19 y+42$ be the given polynomial.
Now, putting $y=2$, we get
$f(2)=2(2)^3-5(2)^2-19(2)+42$
$=16-20-38+42$
$=-58+58=0$
Therefore, $(y-2)$ is a factor of polynomial $f(y)$.
Now,
$f(y)=2 y^2(y-2)-y(y-2)-21(y-2)$
$=(y-2)\left(2 y^2-y-21\right)$
$=(y-2)\left(2 y^2-7 y+6 y-21\right)$
$=(y-2)(y+3)(2 y-7)$
Hence $(y-2),(y+3)$ and $(2 y-7)$ are the factors of polynomial $f(y)$.
View full question & answer→Question 354 Marks
Using factor theorem, factorize the following polynomials:
$x^3 - 10x^2 - 53x - 42$
AnswerLet $f(x)=x^3-10 x^2-53 x-42$ be the given polynomial.
Now, putting $x=-1$, we get
$f(-1)=(-1)^3-10(-1)^2-53(-1)-42$
$=-1-10+53-42$
$=-53+53=0$
Therefore, $(x+1)$ is a factor of polynomial $f(x)$.
Now,
$f(x)=x^2(x+1)-11 x(x+1)-42(x+1)$
$=(x+1)\left(x^2-11 x-42\right)$
$=(x+1)\left(x^2-14 x+3 x-42\right)$
$=(x+1)(x+3)(x-14)$
Hence $(x+1),(x+3)$ and $(x-14)$ are the factors of polynomial $f(x)$.
View full question & answer→Question 364 Marks
If $x=0$ and $x=-1$ are the roots of the polynomial $f(x)=2 x^3-3 x^2+a x+b$, Find the of $a$ and $b$.
AnswerWe know that, $f(x)=2 x^3-3 x^2+a x+b$
Given, the values of $x$ are 0 and -1
Substitute $x=0$ in $f(x)$
$f(0)=2(0)^3-3(0)^2+a(0)+b$
$=0-0+0+b$
$=b \ldots(1)$
Substitute $x =(-1)$ in $f ( x )$
$f(-1)=2(-1)^3-3(-1)^2+a(-1)+b$
$=-2-3-a+b$
$=-5-a+b \ldots(2)$
We need to equate equations 1 and 2 to zero
$b=0 \text { and }-5-a+b=0$
since, the value of $b$ is zero
substitute $b=0$ in equation 2
$\Rightarrow-5-a=-b$
$\Rightarrow-5-a=0$
$a=-5$
the values of $a$ and $b$ are -5 and 0 respectively.
View full question & answer→Question 374 Marks
Using factor theorem, factorize the following polynomials:
$x^3 - 6x^2 + 3x + 10$
AnswerLet $x=2 f(2)=2^3+6(2)^2+3(2)+10=8-24+6+10=0$
$\therefore x=2 \text { is a solution } f(x)$
i. $e(x-2)$ is a factor of $f(x)$
By division algorithm $x^3-6 x^2+3 x+10$
$=(x-2)\left(x^2-4 x-5\right)$
$=(x-2)\left(x^2-5 x+x-5\right)$
$=(x-2)(x(x-5)+1(x-5))$
$=(x-2)(x-5)(x+1)$
$\therefore x^3-6 x^2+3 x+10$
$=(x-2)(x-5)(x+1)$ View full question & answer→Question 384 Marks
Find the integral roots of the polynomial f(x) =$ x^3+ 6x^2 + 11x + 6$
AnswerGiven, that $f(x)=x^3+6 x^2+11 x+6$
Clearly we can say that, the polynomial $f ( x )$ with an integer coefficient and the highest degree term coefficient which is known as leading factor is 1 .
So, the roots of $f(x)$ are limited to integer factor of 6 , they are $\pm 1, \pm 2, \pm 3, \pm 6$
Let $x=-1$
$f(-1)=(-1)^3+6(-1)^2+11(-1)+6$
$=-1+6-11+6$
$=0$
Let $x=-2$
$f(-2)=(-2)^3+6(-2)^2+11(-2)+6$
$=-8-(6 \times 4)-22+6$
$=-8+24-22+6$
$=0$
Let $x=-3$
$f(-3)=(-3)^3+6(-3)^2+11(-3)+6$
$=-27-(6 \times 9)-33+6$
$=-27+54-33+6$
$=0$
But from all the given factors only $-1,-2,-3$ gives the result as zero.
So, the integral multiples of $x^3+6 x^2+11 x+6$ are $-1,-2,-3$.
View full question & answer→