Question 14 Marks
Divide the polynomial $(2y^4 - 3y^3 + 5y - 4) ÷ (y - 1)$ By synthetic division method and Linear method.
View full question & answer→Questions
4 Mark Question
Take a timed test
8 questions · self-marked practice — reveal the answer and mark yourself.
Question 24 Marks
Divide the polynomial $(3x^3 + 2x^2 - 1)$ by $(x + 2)$. By synthetic division method and Linear method.
View full question & answer→Question 34 Marks
Ex (5) Divide $\left(2+2 x^2\right) \div(x+2)$ and write the answer in the given form Dividend $=$ Divisor $\times$ Quotient + Remainder
View full question & answer→Question 44 Marks
i. If $p(x)=2+5 x$, then find the value of $p(2)+p(-2)-p(1)$.
ii. If $p(x)=2 x^2-5 \sqrt{ } 3 x+5$, then find the value of $p(5 \sqrt{ } 3)$.
ii. If $p(x)=2 x^2-5 \sqrt{ } 3 x+5$, then find the value of $p(5 \sqrt{ } 3)$.
Answer
View full question & answer→$\text { i. } p(x)=2+5 x$
$\therefore P(2)=2+5(2)$
$=2+10$
$=12$
$p(x)=2+5 x$
$P(-2)=2+5(-2)$
$=2-10=-8$
$p(x)=2+5 x$
$P(1)=2+5(1)$
$=2+5=7$
$\therefore P(2)+P(-2)-p(1)=12+(-8)-7$
$\therefore P(2)+p(-2)-p(1)=-3$
$\text { ii. } p(x)=2 x^2-5 \sqrt{ } 3 x+5$
$\therefore p(5 \sqrt{ } 3)=2(5 \sqrt{ } 3)^2-5 \sqrt{ } 3(5 \sqrt{ } 3)+5$
$=2(25 \times 3)-25 x 3+5$
$=150-75+5$
$\therefore p(5 \sqrt{ } 3)=80$
$\therefore P(2)=2+5(2)$
$=2+10$
$=12$
$p(x)=2+5 x$
$P(-2)=2+5(-2)$
$=2-10=-8$
$p(x)=2+5 x$
$P(1)=2+5(1)$
$=2+5=7$
$\therefore P(2)+P(-2)-p(1)=12+(-8)-7$
$\therefore P(2)+p(-2)-p(1)=-3$
$\text { ii. } p(x)=2 x^2-5 \sqrt{ } 3 x+5$
$\therefore p(5 \sqrt{ } 3)=2(5 \sqrt{ } 3)^2-5 \sqrt{ } 3(5 \sqrt{ } 3)+5$
$=2(25 \times 3)-25 x 3+5$
$=150-75+5$
$\therefore p(5 \sqrt{ } 3)=80$
Question 54 Marks
By using factor theorem in the following examples, determine whether $q(x)$ is a factor of $p(x)$ or not.
i. $p(x)=x^3-x^2-x-1 ; q(x)=x-1$
ii. $p(x)=2 x^3-x^2-45 ; q(x)=x-3$
i. $p(x)=x^3-x^2-x-1 ; q(x)=x-1$
ii. $p(x)=2 x^3-x^2-45 ; q(x)=x-3$
Answer
View full question & answer→i. $p(x)$ =$ x^3 – x^2 – x – 1$
Divisor =$q(x)$ $= x – 1$
∴ take $x = 1$
Remainder = p(1)
$p(x) = x^3 – x^2 – x – 1$
$\therefore P(1) = (1)^3 – (1)^2 – 1 – 1$
$= 1 – 1 – 1 – 1$
$= -2 \neq 0$
∴ By factor theorem, $x – 1$ is not a factor of$ x^3 – x^2 – x – 1$.ii. $p(x) = 2x^3 – x – 45$
Divisor $= q(x) = x – 3$
take $x = 3$
Remainder =$ p(3)$
$p(x) = 2x^3 – x^2 – 45$
$P(3) = 2(3)^3 – (3)^2 – 45$
$= 2(27) – 9 – 45$
$= 54 – 9 – 45$
$= 0$
$\therefore$ By factor theorem, $x – 3$ is a factor of $2x^3 – x^2 – 45.$
Divisor =$q(x)$ $= x – 1$
∴ take $x = 1$
Remainder = p(1)
$p(x) = x^3 – x^2 – x – 1$
$\therefore P(1) = (1)^3 – (1)^2 – 1 – 1$
$= 1 – 1 – 1 – 1$
$= -2 \neq 0$
∴ By factor theorem, $x – 1$ is not a factor of$ x^3 – x^2 – x – 1$.ii. $p(x) = 2x^3 – x – 45$
Divisor $= q(x) = x – 3$
take $x = 3$
Remainder =$ p(3)$
$p(x) = 2x^3 – x^2 – 45$
$P(3) = 2(3)^3 – (3)^2 – 45$
$= 2(27) – 9 – 45$
$= 54 – 9 – 45$
$= 0$
$\therefore$ By factor theorem, $x – 3$ is a factor of $2x^3 – x^2 – 45.$
Question 64 Marks
Divide first polynomial by second polynomial and write the answer in the form 'Dividend = Divisor x Quotient + Remainder':
$5 x^5+4 x^4-3 x^3+2 x^2+2 ; x^2-x$
$5 x^5+4 x^4-3 x^3+2 x^2+2 ; x^2-x$
Answer
View full question & answer→$5 x^5+4 x^4-3 x^3+2 x^2+2=5 x^5+4 x^4-3 x^3+2 x+0 x+2$
$\therefore$ Quotient $=5 x^3+9 x^2+6 x+8$,
Remainder $=8 x +2$
Now, Dividend = Divisor $x$ Quotient + Remainder
$\therefore 5 x^5+4 x^4-3 x^3+2 x^2+2=\left(x^2-x\right)\left(5 x^3+9 x^2+6 x+8\right)+(8 x+2)$
$\therefore$ Quotient $=5 x^3+9 x^2+6 x+8$,
Remainder $=8 x +2$
Now, Dividend = Divisor $x$ Quotient + Remainder
$\therefore 5 x^5+4 x^4-3 x^3+2 x^2+2=\left(x^2-x\right)\left(5 x^3+9 x^2+6 x+8\right)+(8 x+2)$
Question 74 Marks
Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor x Quotient + Remainder’ :
$x^3 – 64; x – 4$
$x^3 – 64; x – 4$
Answer
View full question & answer→$x^3 – 64 = x3 + 0x^2 + 0x – 64$
$\therefore$ Quotient = $x^2 + 4x + 16$, Remainder = $0$
Now, Dividend = Divisor x Quotient + Remainder
$\therefore x^3 – 64 = (x – 4)(x^2 + 4x + 16) + 0$
$\therefore$ Quotient = $x^2 + 4x + 16$, Remainder = $0$
Now, Dividend = Divisor x Quotient + Remainder
$\therefore x^3 – 64 = (x – 4)(x^2 + 4x + 16) + 0$
Question 84 Marks
Write the following polynomials in coefficient form.
i. $x^3-2$
ii. $5 y$
iii. $2 m^4-3 m^2+7$
iv. $-\frac{2}{3}$
i. $x^3-2$
ii. $5 y$
iii. $2 m^4-3 m^2+7$
iv. $-\frac{2}{3}$
Answer
View full question & answer→i. $x^3-2=x^3+0 x^2+0 x-2$
$\therefore$ Coefficient form of the given polynomial $=(1,0,0,-2)$ ii. $5 y=5 y+0$
$\therefore$ Coefficient form of the given polynomial $=(5,0)$
iii. $2 m^4-3 m^2+7$
$=2 m^4+O m^3-3 m^2+0 m+7$
$\therefore$ Coefficient form of the given polynomial $=(2,0,-3,0,7)$
iv. $-\frac{2}{3}$
$\therefore$ Coefficient form of the given polynomial $=\left(-\frac{2}{3}\right)$
$\therefore$ Coefficient form of the given polynomial $=(1,0,0,-2)$ ii. $5 y=5 y+0$
$\therefore$ Coefficient form of the given polynomial $=(5,0)$
iii. $2 m^4-3 m^2+7$
$=2 m^4+O m^3-3 m^2+0 m+7$
$\therefore$ Coefficient form of the given polynomial $=(2,0,-3,0,7)$
iv. $-\frac{2}{3}$
$\therefore$ Coefficient form of the given polynomial $=\left(-\frac{2}{3}\right)$