Question 12 Marks
Add : $3 m^2 n+5 m n^2-7 m n$ and $2 m^2 n-m n^2+m n$.
View full question & answer→Questions
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Question 22 Marks
Multiply : $\left(m^2-5\right) \times\left(m^3+2 m-2\right)$
View full question & answer→Question 32 Marks
Subtract : $5 a^2-2 a$ from $7 a^2+5 a+6$.
View full question & answer→Question 42 Marks
Find the factors of the polynomials given below : $\frac{1}{2} x^2-3 x+4$
Answer
View full question & answer→$ \frac{1}{2} x^2-3 x+4$
$=\frac{1}{2} x^2-2 x-x+4$
$=\frac{1}{2} x^2-\frac{2 \times 2}{2} x-x+4$
$=\frac{1}{2} x(x-4)-1(x-4)$
$=(x-4)\left(\frac{1}{2} x-1\right)$
Alternate method
$\frac{1}{2} x^2-3 x+4=\frac{1}{2}\left(x^2-6 x+8\right)$
$=\frac{1}{2}\left(x^2-4 x-2 x+8\right)$
$=\frac{1}{2}[x(x-4)-2(x-4)]$
$=\frac{1}{2}(x-2)(x-4)$
$=\frac{1}{2} x^2-2 x-x+4$
$=\frac{1}{2} x^2-\frac{2 \times 2}{2} x-x+4$
$=\frac{1}{2} x(x-4)-1(x-4)$
$=(x-4)\left(\frac{1}{2} x-1\right)$
Alternate method
$\frac{1}{2} x^2-3 x+4=\frac{1}{2}\left(x^2-6 x+8\right)$
$=\frac{1}{2}\left(x^2-4 x-2 x+8\right)$
$=\frac{1}{2}[x(x-4)-2(x-4)]$
$=\frac{1}{2}(x-2)(x-4)$
Question 52 Marks
Find the factors of the polynomials given below: $\sqrt{3} x^2+4 x+\sqrt{3}$
Answer
View full question & answer→$\sqrt{3} \times 2+4 x+\sqrt{3}$
$=\sqrt{3} \times 2+3 x+x+\sqrt{3}$
$=\sqrt{3} \times 2+\sqrt{3} x \sqrt{3} x+x+\sqrt{3}$
$=\sqrt{3} x(x+\sqrt{3})+1(x+\sqrt{3})$
$=(x+\sqrt{3})(\sqrt{3} x+1)$
$=\sqrt{3} \times 2+3 x+x+\sqrt{3}$
$=\sqrt{3} \times 2+\sqrt{3} x \sqrt{3} x+x+\sqrt{3}$
$=\sqrt{3} x(x+\sqrt{3})+1(x+\sqrt{3})$
$=(x+\sqrt{3})(\sqrt{3} x+1)$
Question 62 Marks
Find the factors of the polynomials given below: $3 y^2-2 y-1$
Answer
View full question & answer→$3 y^2-2 y-1$
$=3 y^2-3 y+y-1$
$=3 y(y-1)+1(y-1)$
$=(y-1)(3 y+1)$
$=3 y^2-3 y+y-1$
$=3 y(y-1)+1(y-1)$
$=(y-1)(3 y+1)$
Question 72 Marks
Find the factors of the polynomials given below: $12 x^2+61 x+77$
Answer
View full question & answer→$12 x^2+61 x+77$
$=12 x^2+28 x+33 x+77$
$=4 x(3 x+7) 4+11(3 x+7)$
$=(3 x+7)(4 x+11)$
$=12 x^2+28 x+33 x+77$
$=4 x(3 x+7) 4+11(3 x+7)$
$=(3 x+7)(4 x+11)$
Question 82 Marks
Find the factors of the polynomials given below: $2 m^2+5 m-3$
Answer
View full question & answer→$2 m^2+5 m-3$
$=2 m^2+6 m-m-3$
$=2 m(m+3)-1(m+3)$
$=(m+3)(2 m-1)$
$=2 m^2+6 m-m-3$
$=2 m(m+3)-1(m+3)$
$=(m+3)(2 m-1)$
Question 92 Marks
Find the factors of the polynomials given below: $2 x^2+x-1$
Answer
View full question & answer→$2 x^2+x-1$
$=2 x^2+2 x-x-1$
$=2 x(x+1)-1(x+1)$
$=(x+1)(2 x-1)$
$=2 x^2+2 x-x-1$
$=2 x(x+1)-1(x+1)$
$=(x+1)(2 x-1)$
Question 102 Marks
Factorize the following polynomials: $(x-3)(x-4) 2(x-5)-6$
Answer
View full question & answer→$(x-3)(x-4) 2(x-5)-6$
$=(x-3)(x-5)(x-4)^2-6$
$=\left(x^2-5 x-3 x+15\right)\left(x^2-8 x+16\right)-6$
$=\left(x^2-8 x+15\right)\left(x^2-8 x+16\right)-6$
$=(m+15)(m+16)-6 \ldots\left[\text { Putting } x^2-8 x=m\right]$
$=m(m+16)+15(m+16)-6$
$=m^2+16 m+15 m+240-6$
$=m^2+31 m+234$
$=m^2+18 m+13 m+234$
$=m(m+18)+13(m+18)$
$=(m+18)(m+13)$
$=\left(x^2-8 x+18\right)\left(x^2-8 x+13\right) \ldots\left[\text { Replacing } m=x^2-8 x\right]$
$=(x-3)(x-5)(x-4)^2-6$
$=\left(x^2-5 x-3 x+15\right)\left(x^2-8 x+16\right)-6$
$=\left(x^2-8 x+15\right)\left(x^2-8 x+16\right)-6$
$=(m+15)(m+16)-6 \ldots\left[\text { Putting } x^2-8 x=m\right]$
$=m(m+16)+15(m+16)-6$
$=m^2+16 m+15 m+240-6$
$=m^2+31 m+234$
$=m^2+18 m+13 m+234$
$=m(m+18)+13(m+18)$
$=(m+18)(m+13)$
$=\left(x^2-8 x+18\right)\left(x^2-8 x+13\right) \ldots\left[\text { Replacing } m=x^2-8 x\right]$
Question 112 Marks
Factorize the following polynomials : $\left(y^2+5 y\right)\left(y^2+5 y-2\right)-24$
Answer
View full question & answer→$\left(y^2+5 y\right)\left(y^2+5 y-2\right)-24$
$=(m)(m-2)-24 \ldots\left[\text { Putting } y^2+5 y=m\right]$
$=m^2-2 m-24$
$=m^2-6 m+4 m-24$
$=m(m-6)+4(m-6)$
$=(m-6)(m+4)$
$=\left(y^2+5 y-6\right)\left(y^2+5 y+4\right) \ldots\left[\text { Replacing } m=y^2+5 y\right]$
$=\left(y^2+6 y-y-6\right)\left(y^2+4 y+y+4\right)$
$=[y(y+6)-1(y+6)][y(y+4)+1(y+4)]$
$=(y+6)(y-1)(y+4)(y+1)$
$=(m)(m-2)-24 \ldots\left[\text { Putting } y^2+5 y=m\right]$
$=m^2-2 m-24$
$=m^2-6 m+4 m-24$
$=m(m-6)+4(m-6)$
$=(m-6)(m+4)$
$=\left(y^2+5 y-6\right)\left(y^2+5 y+4\right) \ldots\left[\text { Replacing } m=y^2+5 y\right]$
$=\left(y^2+6 y-y-6\right)\left(y^2+4 y+y+4\right)$
$=[y(y+6)-1(y+6)][y(y+4)+1(y+4)]$
$=(y+6)(y-1)(y+4)(y+1)$
Question 122 Marks
Factorize the following polynomials: $(y+2)(y-3)(y+8)(y+3)+56$
Answer
View full question & answer→$(y+2)(y-3)(y+8)(y+3)+56$
$=(y+2)(y+3)(y-3)(y+8)+56$
$=\left(y^2+3 y+2 y+6\right)\left(y^2+8 y-3 y-24\right)+56$
$=\left(y^2+5 y+6\right)\left(y^2+5 y-24\right)+56$
$=(m+6)(m-24)+56 \ldots[\text { Putting } y 2+5 y=m]$
$=m(m-24)+6(m-24)+56$
$=m^2-24 m+6 m-144+56$
$=m^2-18 m-88$
$=m^2-22 m+4 m-88$
$=m(m-22)+4(m-22)$
$=(m-22)(m+4)$
$=\left(y^2+5 y-22\right)\left(y^2+5 y+4\right) \ldots\left[\text { Replacing } m=y^2+5 y\right]$
$=\left(y^2+5 y-22\right)\left(y^2+4 y+y+4\right)$
$=\left(y^2+5 y-22\right)[y(y+4)+1(y+4)]$
$=\left(y^2+5 y-22\right)(y+4)(y+1)$
$=(y+2)(y+3)(y-3)(y+8)+56$
$=\left(y^2+3 y+2 y+6\right)\left(y^2+8 y-3 y-24\right)+56$
$=\left(y^2+5 y+6\right)\left(y^2+5 y-24\right)+56$
$=(m+6)(m-24)+56 \ldots[\text { Putting } y 2+5 y=m]$
$=m(m-24)+6(m-24)+56$
$=m^2-24 m+6 m-144+56$
$=m^2-18 m-88$
$=m^2-22 m+4 m-88$
$=m(m-22)+4(m-22)$
$=(m-22)(m+4)$
$=\left(y^2+5 y-22\right)\left(y^2+5 y+4\right) \ldots\left[\text { Replacing } m=y^2+5 y\right]$
$=\left(y^2+5 y-22\right)\left(y^2+4 y+y+4\right)$
$=\left(y^2+5 y-22\right)[y(y+4)+1(y+4)]$
$=\left(y^2+5 y-22\right)(y+4)(y+1)$
Question 132 Marks
Factorize the following polynomials: $\left(x^2-2 x+3\right)\left(x^2-2 x+5\right)-35$
Answer
$\left(x^2-2 x+3\right)\left(x^2-2 x+5\right)-35$
$=(m+3)(m+5)-35 \ldots\left[\text { Putting } x^2-2 x=m\right]$
$=m(m+5)+3(m+5)-35$
$=m^2+5 m+3 m+15-35$
$=m^2+8 m-20$
$=m^2+10 m-2 m-20$
$=m(m+10)-2(m+10)$
$=(m+10)(m-2)$
$=\left(x^2-2 x+10\right)\left(x^2-2 x-2\right) \ldots\left[\right.$ Replacing $\left.m=x^2-2 x\right]$
$\overbrace{10}^{-20}$
$10 \times-2=-20$
$10-2=8$
View full question & answer→$\left(x^2-2 x+3\right)\left(x^2-2 x+5\right)-35$
$=(m+3)(m+5)-35 \ldots\left[\text { Putting } x^2-2 x=m\right]$
$=m(m+5)+3(m+5)-35$
$=m^2+5 m+3 m+15-35$
$=m^2+8 m-20$
$=m^2+10 m-2 m-20$
$=m(m+10)-2(m+10)$
$=(m+10)(m-2)$
$=\left(x^2-2 x+10\right)\left(x^2-2 x-2\right) \ldots\left[\right.$ Replacing $\left.m=x^2-2 x\right]$
$\overbrace{10}^{-20}$
$10 \times-2=-20$
$10-2=8$
Question 142 Marks
Factorize the following polynomials: $\left(x^2-6 x\right)^2-8\left(x^2-6 x+8\right)-64$
Answer
View full question & answer→$\left(x^2-6 x\right)^2-8\left(x^2-6 x+8\right)-64$
$=m^2-8(m+8)-64 \ldots\left[\text { Putting } x^2-6 x=m\right]$
$=m^2-8 m-64-64$
$=m^2-8 m-128$
$=m^2-16 m+8 m-128$
$=m(m-16)+8(m-16)$
$=(m-16)(m+8)$
$=\left(x^2-6 x-16\right)\left(x^2-6 x+8\right) \ldots\left[\text { Replacing } m=x^2-6 x\right]$
$=\left(x^2-8 x+2 x-16\right)\left(x^2-4 x-2 x+8\right)$
$=[x(x-8)+2(x-8)][x(x-4)-2(x-4)]$
$=(x-8)(x+2)(x-4)(x-2)$
$=m^2-8(m+8)-64 \ldots\left[\text { Putting } x^2-6 x=m\right]$
$=m^2-8 m-64-64$
$=m^2-8 m-128$
$=m^2-16 m+8 m-128$
$=m(m-16)+8(m-16)$
$=(m-16)(m+8)$
$=\left(x^2-6 x-16\right)\left(x^2-6 x+8\right) \ldots\left[\text { Replacing } m=x^2-6 x\right]$
$=\left(x^2-8 x+2 x-16\right)\left(x^2-4 x-2 x+8\right)$
$=[x(x-8)+2(x-8)][x(x-4)-2(x-4)]$
$=(x-8)(x+2)(x-4)(x-2)$
Question 152 Marks
Factorize the following polynomials: $(x-5)^2-(5 x-25)-24$
Answer
View full question & answer→$(x-5)^2-(5 x-25)-24$
$=(x-5)^2-(5 x-25)-24$
$=(x-5)^2-5(x-5)-24$
$=m^2-5 m-24 \ldots[\text { Putting } x-5=m \text { ] }$
$=m^2-8 m+3 m-24$
$=m(m-8)+3(m-8)$
$=(m-8)(m+3)$
$=(x-5-8)(x-5+3) \ldots \text { [Replacing } m=x-5]$
$=(x-13)(x-2)$
$=(x-5)^2-(5 x-25)-24$
$=(x-5)^2-5(x-5)-24$
$=m^2-5 m-24 \ldots[\text { Putting } x-5=m \text { ] }$
$=m^2-8 m+3 m-24$
$=m(m-8)+3(m-8)$
$=(m-8)(m+3)$
$=(x-5-8)(x-5+3) \ldots \text { [Replacing } m=x-5]$
$=(x-13)(x-2)$
Question 162 Marks
Factorize the following polynomials: $\left(x^2-x\right)^2-8\left(x^2-x\right)+12$
Answer
View full question & answer→$\left(x^2-x\right)^2-8\left(x^2-x\right)+12$
$=m^2-8 m+12 \ldots\left[\text { Putting } x^2-x=m\right]$
$=m^2-6 m-2 m+12$
$=m(m-6)-2(m-6)$
$=(m-6)(m-2)$
$=\left(x^2-x-6\right)\left(x^2-x-2\right) \ldots\left[\text { Replacing } m=x^2-x\right]$
$=\left(x^2-3 x+2 x-6\right)\left(x^2-2 x+x-2\right)$
$=[x(x-3)+2(x-3)][x(x-2)+1(x-2)]$
$=(x-3)(x+2)(x-2)(x+1)$
$=m^2-8 m+12 \ldots\left[\text { Putting } x^2-x=m\right]$
$=m^2-6 m-2 m+12$
$=m(m-6)-2(m-6)$
$=(m-6)(m-2)$
$=\left(x^2-x-6\right)\left(x^2-x-2\right) \ldots\left[\text { Replacing } m=x^2-x\right]$
$=\left(x^2-3 x+2 x-6\right)\left(x^2-2 x+x-2\right)$
$=[x(x-3)+2(x-3)][x(x-2)+1(x-2)]$
$=(x-3)(x+2)(x-2)(x+1)$
Question 172 Marks
Verify that $(x-1)$ is a factor of the polynomial $x^3+4 x-5$.
Answer
View full question & answer→Here, $p ( x )= x ^3+4 x -5$
Substituting $x=1$ in $p(x)$, we get
$p(1)=(1)^3+4(1)-5$
$=1+4-5$
$P(1)=0$
$\therefore$ By remainder theorem,
Remainder $=0$
$\therefore(x-1)$ is the factor of $x ^3+4 x -5$.
Substituting $x=1$ in $p(x)$, we get
$p(1)=(1)^3+4(1)-5$
$=1+4-5$
$P(1)=0$
$\therefore$ By remainder theorem,
Remainder $=0$
$\therefore(x-1)$ is the factor of $x ^3+4 x -5$.
Question 182 Marks
If $\left(x^{31}+31\right)$ is divided by $(x+1)$, then find the remainder.
Answer
View full question & answer→$p(x)=x^{31}+31$
$\text { Divisor }=x+1$
$\therefore \text { take } x=-1$
$\therefore \text { By remainder theorem, }$
$\text { Remainder }=p(-1)$
$p(x)=x^{31}+31 \ldots$
$\therefore p(-1)=(-1)^{31}+31$
$=-1+31=30$
$\therefore \text { Remainder }=30$
$\text { Divisor }=x+1$
$\therefore \text { take } x=-1$
$\therefore \text { By remainder theorem, }$
$\text { Remainder }=p(-1)$
$p(x)=x^{31}+31 \ldots$
$\therefore p(-1)=(-1)^{31}+31$
$=-1+31=30$
$\therefore \text { Remainder }=30$
Question 192 Marks
Divide the first polynomial by the second polynomial and find the remainder using remainder theorem :
$\left(54 m^3+18 m^2-27 m+5\right) ;(m-3)$
$\left(54 m^3+18 m^2-27 m+5\right) ;(m-3)$
Answer
View full question & answer→$p(m)=54 m^3+18 m^2-27 m+5$
Divisor $=m-3$
$\therefore$ take $m =3$
$\therefore$ By remainder theorem,
Remainder $=p(3)$
$p(m)=54 m^3+18 m^2-27 m+5$
$\therefore p(3)=54(3)^3+18(3)^2-27(3)+5$
$=54(27)+18(9)-81+5$
$=1458+162-81+5$
$\therefore \text { Remainder }=1544$
Divisor $=m-3$
$\therefore$ take $m =3$
$\therefore$ By remainder theorem,
Remainder $=p(3)$
$p(m)=54 m^3+18 m^2-27 m+5$
$\therefore p(3)=54(3)^3+18(3)^2-27(3)+5$
$=54(27)+18(9)-81+5$
$=1458+162-81+5$
$\therefore \text { Remainder }=1544$
Question 202 Marks
Divide the first polynomial by the second polynomial and find the remainder using remainder theorem :
$\left(2 x^3-2 x^2+a x-a\right) ;(x-a)$
$\left(2 x^3-2 x^2+a x-a\right) ;(x-a)$
Answer
View full question & answer→$p(x)=2 x^3-2 x^2+a x-a$
Divisor $=x-a$
$\therefore$ take $x = a$
By remainder theorem,
Remainder $=p(a)$
$p(x)=2 x^3-2 x^2+a x-a$
$\therefore p ( a )=2 a ^3-2 a ^2+ a ( a )- a$
$=2 a^3-2 a^2+a^2-a$
$\therefore$ Remainder $=2 a^3-a^2-a$
Divisor $=x-a$
$\therefore$ take $x = a$
By remainder theorem,
Remainder $=p(a)$
$p(x)=2 x^3-2 x^2+a x-a$
$\therefore p ( a )=2 a ^3-2 a ^2+ a ( a )- a$
$=2 a^3-2 a^2+a^2-a$
$\therefore$ Remainder $=2 a^3-a^2-a$
Question 212 Marks
Divide the first polynomial by the second polynomial and find the remainder using remainder theorem :
$(x^2 – 1x + 9); (x + 1)$
$(x^2 – 1x + 9); (x + 1)$
Answer
View full question & answer→p(x) = $x^2 – 7x + 9$
Divisor =$x + 1$
\therefore take $x = – 1$
\therefore By remainder theorem,
\therefore Remainder =$p(-1)$
p(x) = $x^2 – 7x + 9$
$\therefore p(-1) = (- 1)^2 – 7(- 1) + 9$
$= 1 + 7 + 9$
$\therefore Remainder =17$
Divisor =$x + 1$
\therefore take $x = – 1$
\therefore By remainder theorem,
\therefore Remainder =$p(-1)$
p(x) = $x^2 – 7x + 9$
$\therefore p(-1) = (- 1)^2 – 7(- 1) + 9$
$= 1 + 7 + 9$
$\therefore Remainder =17$
Question 222 Marks
For the polynomial $mx^2 – 2x + 3$ if $p(-1) = 7$, then find m.
Answer
View full question & answer→$p(x) = mx^2 – 2x + 3$
$\therefore p(- 1) = m (- 1)^2 – 2(- 1) + 3$
$\therefore 7 = m(1) + 2 + 3 …[\because p(-1) = 7]$
$\therefore 7 = m + 5$
$\therefore m = 7 – 5$
$\therefore m = 2$
$\therefore p(- 1) = m (- 1)^2 – 2(- 1) + 3$
$\therefore 7 = m(1) + 2 + 3 …[\because p(-1) = 7]$
$\therefore 7 = m + 5$
$\therefore m = 7 – 5$
$\therefore m = 2$
Question 232 Marks
If the value of the polynomial $m^3 + 2m + a$ is $12$ for $m = 2$, then find the value of a.$F$
Answer
View full question & answer→$p(m) = m^3 + 2m + a$
$\therefore p(2) = (2)^3 + 2(2) + a$
$\therefore 12 = 8 + 4 + a … [\because p(2)= 12]$
$\therefore 12 = 12 + a$
$\therefore a = 12 – 12$
$\therefore a = 0$
$\therefore p(2) = (2)^3 + 2(2) + a$
$\therefore 12 = 8 + 4 + a … [\because p(2)= 12]$
$\therefore 12 = 12 + a$
$\therefore a = 12 – 12$
$\therefore a = 0$
Question 242 Marks
For each of the following polynomial, find $p(1), p(0)$ and $p(-2): p(y)=x^4-2 x^2+x$
Answer
View full question & answer→$p(x)=x^4-2 x^2-x$
$\therefore p(1)=(1)^4-2(1)^2-1$
$=1-2-1$
$\therefore p(1)=-2$
$\therefore p(x)=x^4-2 x^2-x$
$\therefore p(0)=(0)^4-2(0)^2-0$
$=0-0-0$
$\therefore p(0)=0$
$p(x)=x^4-2 x^2-x$
$\therefore p(-2)=(-2)^4-2(-2)^2-(-2)$
$=16-2(4)+2$
$=16-8+2$
$\therefore p(-2)=10$
$\therefore p(1)=(1)^4-2(1)^2-1$
$=1-2-1$
$\therefore p(1)=-2$
$\therefore p(x)=x^4-2 x^2-x$
$\therefore p(0)=(0)^4-2(0)^2-0$
$=0-0-0$
$\therefore p(0)=0$
$p(x)=x^4-2 x^2-x$
$\therefore p(-2)=(-2)^4-2(-2)^2-(-2)$
$=16-2(4)+2$
$=16-8+2$
$\therefore p(-2)=10$
Question 252 Marks
For each of the following polynomial, find $p(1), p(0)$ and $p(-2): p(y)=y^2-2 y+5$
Answer
View full question & answer→$p(y)=y^2-2 y+5$
$\therefore p(1)=1^2-2(1)+5$
$=1-2+5$
$\therefore P(1)=4$
$p(y)=y^2-2 y+5$
$\therefore p(0)=0^2-2(0)+5$
$=0-0+5$
$\therefore p(0)=5$
$p(y)=y^2-2 y+5$
$\therefore p(-2)=(-2)^2-2(-2)+5$
$=4+4+5$
$\therefore p(-2)=13$
$\therefore p(1)=1^2-2(1)+5$
$=1-2+5$
$\therefore P(1)=4$
$p(y)=y^2-2 y+5$
$\therefore p(0)=0^2-2(0)+5$
$=0-0+5$
$\therefore p(0)=5$
$p(y)=y^2-2 y+5$
$\therefore p(-2)=(-2)^2-2(-2)+5$
$=4+4+5$
$\therefore p(-2)=13$
Question 262 Marks
For each of the following polynomial, find $p(1), p(0)$ and $p(-2): p(x)=x^3$
Answer
View full question & answer→$p(x)=x^3$
$\therefore p(1)=1^3=1$
$p(x)=x^3$
$\therefore p(0)=0^3=0$
$p(x)=x^3$
$\therefore p(-2)=(-2)^3=-8$
$\therefore p(1)=1^3=1$
$p(x)=x^3$
$\therefore p(0)=0^3=0$
$p(x)=x^3$
$\therefore p(-2)=(-2)^3=-8$
Question 272 Marks
If $p(y)=2 y^3-6 y^2-5 y+7$, then find $p(2)$.
Answer
View full question & answer→$p(y)=2 y^3-6 y^2-5 y+7$
Put $y=2$ in the given polynomial.
$\therefore p(2)=2(2)^3-6(2)^2-5(2)+7$
$=2 \times 8-6 \times 4-10+7$
$=16-24-10+7$
$\therefore P(2)=-11$
Put $y=2$ in the given polynomial.
$\therefore p(2)=2(2)^3-6(2)^2-5(2)+7$
$=2 \times 8-6 \times 4-10+7$
$=16-24-10+7$
$\therefore P(2)=-11$
Question 282 Marks
If $p(y) = y^2 – 3√2 + 1,$ then find $p( 3√2 ).$
Answer
View full question & answer→$p(y) = y^2 – 3√2 y + 1$
Putp= $3√2$ in the given polynomial.
$\therefore p( 3√2 ) = (3√2 )^2 – 3√2 (3√2 ) + 1$
$= 9 x 2 – 9 x 2 + 1$
$= 18 – 18 + 1$
$\therefore p( 3√2 ) = 1$
Putp= $3√2$ in the given polynomial.
$\therefore p( 3√2 ) = (3√2 )^2 – 3√2 (3√2 ) + 1$
$= 9 x 2 – 9 x 2 + 1$
$= 18 – 18 + 1$
$\therefore p( 3√2 ) = 1$
Question 292 Marks
For $x=0$, find the value of the polynomial $x^2-5 x+5$.
Answer
View full question & answer→$p(x)=x^2-5 x+5$
Put $x=0$ in the given polynomial.
$\therefore P(0)=(0)^2-5(0)+5$
$=0-0+5$
$\therefore p(0)=5$
Put $x=0$ in the given polynomial.
$\therefore P(0)=(0)^2-5(0)+5$
$=0-0+5$
$\therefore p(0)=5$
Question 302 Marks
Multiply the given polynomials :
$2 y+1 ; y^2-2 y+3 y$
$2 y+1 ; y^2-2 y+3 y$
Answer
View full question & answer→$(2 y+1) \times\left(y^2-2 y^3+3 y\right)$
$=2 y\left(y^2-2 y^3+3 y\right)+1\left(y^2-2 y^3+3 y\right)$
$=2 y^3-4 y^4+6 y^2+y^2-2 y^3+3 y$
$=-4 y^4+2 y^3-2 y^3+6 y^2+y^2+3 y$
$=-4 y^4+7 y^2+3 y$
$=2 y\left(y^2-2 y^3+3 y\right)+1\left(y^2-2 y^3+3 y\right)$
$=2 y^3-4 y^4+6 y^2+y^2-2 y^3+3 y$
$=-4 y^4+2 y^3-2 y^3+6 y^2+y^2+3 y$
$=-4 y^4+7 y^2+3 y$
Question 312 Marks
Multiply the given polynomials :
$x^5-1 ; x^3+2 x^2+2$
$x^5-1 ; x^3+2 x^2+2$
Answer
View full question & answer→$\left(x^5-1\right) \times\left(x^3+2 x^2+2\right)$
$=x^5\left(x^3+2 x^2+2\right)-1\left(x^3+2 x^2+2\right)$
$=x^8+2 x^7+2 x^5-x^3-2 x^2-2$
$=x^5\left(x^3+2 x^2+2\right)-1\left(x^3+2 x^2+2\right)$
$=x^8+2 x^7+2 x^5-x^3-2 x^2-2$
Question 322 Marks
Multiply the given polynomials :
$2 x ; x^2-2 x-1$
$2 x ; x^2-2 x-1$
Answer
View full question & answer→$(2 x) x\left(x^2-2 x-1\right)=2 x^3-4 x^2-2 x$
Question 332 Marks
Subtract the second polynomial from the first :
$2 a b^2+3 a^2 b-4 a b ; 3 a b-8 a b^2+2 a^2 b$
$2 a b^2+3 a^2 b-4 a b ; 3 a b-8 a b^2+2 a^2 b$
Answer
View full question & answer→$\left(2 a b^2+3 a^2 b-4 a b\right)-\left(3 a b-8 a b^2+2 a^2 b\right)$
$=2 a b^2+3 a^2 b-4 a b-3 a b+8 a b^2-2 a^2 b$
$=2 a b^2+8 a b^2+3 a^2 b-2 a^2 b-4 a b-3 a b$
$=10 a b^2+a^2 b-7 a b$
$=2 a b^2+3 a^2 b-4 a b-3 a b+8 a b^2-2 a^2 b$
$=2 a b^2+8 a b^2+3 a^2 b-2 a^2 b-4 a b-3 a b$
$=10 a b^2+a^2 b-7 a b$
Question 342 Marks
Subtract the second polynomial from the first :
$x^2-9 x+\sqrt{3} ;-19 x+\sqrt{3}+7 x^2$
$x^2-9 x+\sqrt{3} ;-19 x+\sqrt{3}+7 x^2$
Answer
View full question & answer→$x^2-9 x+\sqrt{3}-\left(-19 x+\sqrt{3}+7 x^2\right)$
$=x^2-9 x+\sqrt{3}+19 x-\sqrt{3}-7 x^2$
$=x^2-7 x^2-9 x+19 x+\sqrt{3}-\sqrt{3}$
$=-6 x^2+10 x$
$=x^2-9 x+\sqrt{3}+19 x-\sqrt{3}-7 x^2$
$=x^2-7 x^2-9 x+19 x+\sqrt{3}-\sqrt{3}$
$=-6 x^2+10 x$
Question 352 Marks
Add the given polynomials : $2 y^2+7 y+5 ; 3 y+9 ; 3 y^2-4 y-3$
Answer
View full question & answer→$\left(2 y^2+7 y+5\right)+(3 y+9)+\left(3 y^2-4 y-3\right)$
$=2 y^2+7 y+5+3 y+9+3 y^2-4 y-3$
$=2 y^2+3 y^2+7 y+3 y-4 y+5+9-3$
$=5 y^2+6 y+11$
$=2 y^2+7 y+5+3 y+9+3 y^2-4 y-3$
$=2 y^2+3 y^2+7 y+3 y-4 y+5+9-3$
$=5 y^2+6 y+11$
Question 362 Marks
Add the given polynomials : $-7m^4+ 5m^3 + √2 ; 5m^4 – 3m^3 + 2m^2 + 3m – 6$
Answer
View full question & answer→$(-7m^4 + 5m^3 + √2 ) + (5m^4 – 3m^3 + 2m^2 + 3m – 6)$
$= -7m^4 + 5m^3 + √2 + 5m^4 – 3m^3 + 2m^2 + 3m – 6$
$= -7m^4 + 5m^4 + 5m^3 – 3m^3 + 2m^2 + 3m +√2 – 6$
$= -2m^4 + 2m^3 + 2m^2 + 3m + √2 – 6$
$= -7m^4 + 5m^3 + √2 + 5m^4 – 3m^3 + 2m^2 + 3m – 6$
$= -7m^4 + 5m^4 + 5m^3 – 3m^3 + 2m^2 + 3m +√2 – 6$
$= -2m^4 + 2m^3 + 2m^2 + 3m + √2 – 6$
Question 372 Marks
Add the given polynomials : $x^3 – 2x^2 – 9; 5x^3 + 2x + 9$
Answer
View full question & answer→$(x^3 – 2x^2 – 9) + (5x^3 + 2x + 9)$
$= x^3 – 2x^2 – 9 + 5x^3 + 2x + 9$
$= x^3 + 5x^3 – 2x^2 + 2x – 9 + 9$
$= 6x^3 – 2x^2 + 2x$
$= x^3 – 2x^2 – 9 + 5x^3 + 2x + 9$
$= x^3 + 5x^3 – 2x^2 + 2x – 9 + 9$
$= 6x^3 – 2x^2 + 2x$
Question 382 Marks
Write the following polynomials in standard form.
i. $m^3+3+5 m$
ii. $-7 y+y^5+3 y^3-\frac{1}{2}+2 y^4-y^2$
i. $m^3+3+5 m$
ii. $-7 y+y^5+3 y^3-\frac{1}{2}+2 y^4-y^2$
Answer
View full question & answer→i. $m^3+5 m+3$
ii. $y^5+2 y^4+3 y^3-y^2-7 y-\frac{1}{2}$
ii. $y^5+2 y^4+3 y^3-y^2-7 y-\frac{1}{2}$
Question 392 Marks
Classify the following polynomials as linear, quadratic and cubic polynomial.
i. $2 x^2+3 x+1$
ii. $5 p$
iii. $\sqrt{ } 2-\frac{1}{2}$
iv. $m^3+7 m^2+\sqrt[5]{2} m-\sqrt{7}$
V. $a^2$
vi. $3 r^3$
i. $2 x^2+3 x+1$
ii. $5 p$
iii. $\sqrt{ } 2-\frac{1}{2}$
iv. $m^3+7 m^2+\sqrt[5]{2} m-\sqrt{7}$
V. $a^2$
vi. $3 r^3$
Answer
View full question & answer→Linear polynomials: ii, iii
Quadratic polynomials: i, v
Cubic polynomials: iv, vi
Quadratic polynomials: i, v
Cubic polynomials: iv, vi