Solutions of Exercise 2.6 Class 10 Complete Guide
In this post, we’ll explore the solutions to Exercise 2.6 Class 10 Math, where we dive into the efficient technique of synthetic division. This method allows us to simplify polynomial division, particularly when dealing with linear divisors. Synthetic division is not only useful for finding the quotient and remainder of polynomial expressions but also helps in solving equations and finding unknowns when zeros or factors are given. In this exercise 2.6 class 10, we’ll walk through a variety of problems, including finding unknowns, solving cubic and biquadratic equations, and verifying polynomial factors, all using synthetic division. Let’s get started with step-by-step solutions to deepen your understanding of this powerful tool in polynomial algebra.
Question 1 of Exercise 2.6 Class 10. \(\textbf{Use synthetic division to find the quotient and the remainder, when:}\)
(i) \(\boldsymbol{(x^2 + 7x – 1) \div (x + 1)}\)
Since \(x + 1 = 0 \Rightarrow x = -1\)
\[
\begin{array}{r|rrr}
-1 & 1 & 7 & -1 \\
& & -1 & -6 \\
\hline
& 1 & 6 & -7 \\
\end{array}
\]
Quotient: \( Q(x) = x + 6 \)
Remainder: \( R = -7 \)
(ii) \(\boldsymbol{(4x^3 – 5x + 15) \div (x + 3)}\)
Since \(x + 3 = 0 \Rightarrow x = -3\)
\[
\begin{array}{r|rrrr}
-3 & 4 & 0 & -5 & 15 \\
& & -12 & 36 & -93 \\
\hline
& 4 & -12 & 31 & -78 \\
\end{array}
\]
Quotient: \( Q(x) = 4x^2 – 12x + 31 \)
Remainder: \( R = -78 \)
(iii) \(\boldsymbol{(x^3 + x^2 – 3x + 2) \div (x – 2)}\)
Since \(x – 2 = 0 \Rightarrow x = 2\)
\[
\begin{array}{r|rrrr}
2 & 1 & 1 & -3 & 2 \\
& & 2 & 6 & 6 \\
\hline
& 1 & 3 & 3 & 8 \\
\end{array}
\]
Quotient: \( Q(x) = x^2 + 3x + 3 \)
Remainder: \( R = 8 \)
Question 2 of Exercise 2.6 Class 10. \(\textbf{Find the value of (h) using synthetic division:}\)
(i) Find \(h\) if 3 is the zero of the polynomial \(2x^3 – 3hx^2 + 9\)
Let \(P(x) = 2x^3 – 3hx^2 + 0x + 9\) and its zero is 3.
\[
\begin{array}{r|rrrr}
3 & 2 & -3h & 0 & 9 \\
& & 6 & 18-9h & 54-27h \\
\hline
& 2 & 6 & 18 – 9h & 63 – 27h \\
\end{array}
\]
Since remainder = 0, \(63 – 27h = 0\)
\[
h = \frac{63}{27} = \frac{7}{3}
\]
(ii) Find \(h\) if 1 is the zero of the polynomial \(x^3 – 2hx^2 + 11\)
Let \(P(x) = x^3 – 2hx^2 + 0x + 11\) and its zero is 1.
\[
\begin{array}{r|rrrr}
1 & 1 & -2h & 0 & 11 \\
& & 1 & 1 – 2h & 1 – 2h \\
\hline
& 1 & 1 – 2h & 1 – 2h & 12 – 2h \\
\end{array}
\]
Since remainder = 0, \(12 – 2h = 0\)
\[
h = \frac{12}{2} = 6
\]
\(\textbf{(iii) Find the value of \( h \) using synthetic division}\)
\(\textbf{if -1 is the zero of the Polynomial} 2x^3 + 5hx – 23:\)
Let \( P(x) = 2x^3 + 0x^2 + 5hx – 23 \).
Using synthetic division with -1 as the zero:
\[
\begin{array}{r|rrrr}
-1 & 2 & 0 & 5h & -23 \\
& & -2 & 2 & -5h-2 \\
\hline
& 2 & -2 & 5h + 2 & -5h – 25 \\
\end{array}
\]
Since remainder = 0,
\[
-5h – 25 = 0 \quad \Rightarrow \quad -5h = 25 \quad \Rightarrow \quad h = -5
\]
Question 3 of Exercise 2.6 Class 10. \(\textbf{Use synthetic division to find the values of ( l ) and ( m )}.\)
\(\textbf{(i) If } (x + 3) \textbf{ and } (x – 2) \textbf{ are the factors of the polynomial }\)
\(x^3 + 4x^2 + 2lx + m:\)
Since \( x + 3 \) and \( x – 2 \) are factors of \( P(x) = x^3 + 4x^2 + 2lx + m \),
we can apply synthetic division:
For \( x + 3 \) (implies \( x = -3 \)):
\[
\begin{array}{r|rrrr}
-3 & 1 & 4 & 2l & m \\
& & -3 & -3 & -6l + 9 \\
\hline
& 1 & 1 & 2l – 3 & m + 6l – 9 \\
\end{array}
\]
Since remainder = 0:
\[
m – 6l + 9 = 0 \quad \Rightarrow \quad m – 6l = -9 \quad \ldots (i)
\]
For \( x – 2 \) (implies \( x = 2 \)):
\[
\begin{array}{r|rrrr}
2 & 1 & 4 & 2l & m \\
& & 2 & 12 & 4l + 24 \\
\hline
& 1 & 6 & 2l + 12 & m + 4l + 24 \\
\end{array}
\]
Since remainder = 0:
\[
m + 4l + 24 = 0 \quad \Rightarrow \quad m + 4l = -24 \quad \ldots (ii)
\]
Subtracting (ii) from (i):
\[
m – 6l – (m + 4l) = -9 – (-24) \quad \Rightarrow \quad -10l = 15 \quad \Rightarrow \quad l = \frac{-3}{2}
\]
Substitute \( l = \frac{-3}{2} \) into (i):
\[
m – 6\left(\frac{-3}{2}\right) = -9 \quad \Rightarrow \quad m + 9 = -9 \quad \Rightarrow \quad m = -18
\]
\(\textbf{(ii) Find the values of } l \textbf{ and } m \textbf{ if } (x – 1) \textbf{ and } (x + 1)\)
\(\textbf{ are the factors of the polynomial } x^3 – 3lx^2 + 2mx + 6:\)
Given that \( (x – 1) \) and \( (x + 1) \) are factors of the polynomial \( P(x) = x^3 – 3lx^2 + 2mx + 6 \),
the zeros are \( x = 1 \) and \( x = -1 \).
\(\textbf{Step 1: Synthetic division for } x = 1:\)
\[
\begin{array}{r|rrrr}
1 & 1 & -3l & 2m & 6 \\
& & 1 & 1 – 3l & 2m + 1 – 3l \\
\hline
& 1 & 1 & 1 – 3l + 2m & 7 – 3l + 2m \\
\end{array}
\]
Since \( x = 1 \) is a zero, the remainder must be 0:
\[
7 – 3l + 2m = 0 \quad \Rightarrow \quad 3l – 2m = 7 \quad \ldots (i)
\]
\(\textbf{Step 2: Synthetic division for } x = -1:\)
\[
\begin{array}{r|rrrr}
-1 & 1 & -3l & 2m & 6 \\
& & -1 & 3l – 1 & -2m – 3l + 1 \\
\hline
& 1 & -4l & 2m + 4l & 8 \\
\end{array}
\]
Since \( x = -1 \) is a zero, the remainder must be 0:
\[
8 + 2m + 4l = 0 \quad \Rightarrow \quad 2m + 4l = -8 \quad \ldots (ii)
\]
\(\textbf{Step 3: Solving the system of equations}:\)
From equation (ii), rearrange:
\[
2m + 4l = -8 \quad \Rightarrow \quad m + 2l = -4
\]
Substitute \( m = -\frac{1}{2} \) into the first equation:
\[
2\left(-\frac{1}{2}\right) – 3l = -7
\]
Simplifies to:
\[
-1 – 3l = -7 \quad \Rightarrow \quad -3l = -6 \quad \Rightarrow \quad l = 2
\]
Thus, the final values are:
\[
l = 2, \quad m = -\frac{1}{2}
\]
Question 4 of Exercise 2.6 Class 10. \(\textbf{Solve by using synthetic division}.\)
\(\textbf{(i) If 2 is the root of the equation } x^3 – 28x + 48 = 0:\)
Let \( P(x) = x^3 – 28x + 48 \). Since 2 is the root, synthetic division gives:
\[
\begin{array}{r|rrrr}
2 & 1 & 0 & -28 & 48 \\
& & 2 & 4 & -48 \\
\hline
& 1 & 2 & -24 & 0 \\
\end{array}
\]
The depressed equation is:
\[
x^2 + 2x – 24 = 0
\]
Factoring:
\[
(x + 6)(x – 4) = 0 \quad \Rightarrow \quad x = -6, \quad x = 4
\]
Thus, the roots of the given equation are \( x = 2, -6, 4 \).
\(\textbf{(ii) If } 3 \textbf{ is the root of the equation } 2x^3 – 3x^2 – 11x + 6 = 0:\)
\(\textbf{Solution:}\)
Since \( 3 \) is the root of the equation, we perform synthetic division of \( 2x^3 – 3x^2 – 11x + 6 \) by \( x – 3 \):
\[
\begin{array}{r|rrrr}
3 & 2 & -3 & -11 & 6 \\
& & 6 & 9 & -6 \\
\hline
& 2 & 3 & -2 & 0 \\
\end{array}
\]
The depressed equation is:
\[
2x^2 + 3x – 2 = 0
\]
Factoring the quadratic equation:
\[
2x^2 + 4x – x – 2 = 0
\]
\[
2x(x + 2) – 1(x + 2) = 0
\]
\[
(x + 2)(2x – 1) = 0
\]
Therefore, the roots are:
\[
x = -2 \quad \text{or} \quad 2x – 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2}
\]
Thus, the roots of the given equation are:
\[
x = 3, \quad x = -2, \quad \text{and} \quad x = \frac{1}{2}
\]
\(\textbf{(iii) If } -1 \textbf{ is the root of the equation } 4x^3 – x^2 – 11x – 6 = 0:\)
\(\textbf{Solution:}\)
Since \( -1 \) is the root of the equation, we perform synthetic division of \( 4x^3 – x^2 – 11x – 6 \) by \( x + 1 \):
\[
\begin{array}{r|rrrr}
-1 & 4 & -1 & -11 & -6 \\
& & -4 & 5 & 6 \\
\hline
& 4 & -5 & -6 & 0 \\
\end{array}
\]
The depressed equation is:
\[
4x^2 – 5x – 6 = 0
\]
Factoring the quadratic equation:
\[
4x^2 – 8x + 3x – 6 = 0
\]
\[
4x(x – 2) + 3(x – 2) = 0
\]
\[
(x – 2)(4x + 3) = 0
\]
Therefore, the roots are:
\[
x = 2 \quad \text{or} \quad 4x + 3 = 0 \quad \Rightarrow \quad x = \frac{-3}{4}
\]
Thus, the roots of the given equation are:
\[
x = -1, \quad x = 2, \quad \text{and} \quad x = \frac{-3}{4}
\]
Question 5 of Exercise 2.6 Class 10. \(\textbf{Solve by using synthetic division}.\)
\(\textbf{(i) Solve if 1 and 3 are the roots of the equation } x^4 – 10x^2 + 9 = 0:\)
Since 1 and 3 are the roots, synthetic division gives:
For \( x = 1 \):
\[
\begin{array}{r|rrrrr}
1 & 1 & 0 & -10 & 0 & 9 \\
& & 1 & 1 & -9 & -9 \\
\hline
& 1 & 1 & -9 & -9 & 0 \\
\end{array}
\]
For \( x = 3 \):
\[
\begin{array}{r|rrrrr}
3 & 1 & 1 & -9 & -9 \\
& & 3 & 12 & 9 \\
\hline
& 1 & 4 & 3 & 0 \\
\end{array}
\]
The depressed equation is:
\[
x^2 + 4x + 3 = 0 \quad \Rightarrow \quad (x + 3)(x + 1) = 0 \quad \Rightarrow \quad x = -3, -1
\]
Thus, the roots are \( x = 1, 3, -3, -1 \).
\(\textbf{(ii) Solve by using synthetic division, if 3 and -4 are the roots of the equation }\) \(\boldsymbol{x^4 + 2x^3 – 13x^2 – 14x + 24 = 0:}\)
\textbf{Solution:}
Since 3 and -4 are the roots of the equation \( x^4 + 2x^3 – 13x^2 – 14x + 24 = 0 \), we perform synthetic division:
Synthetic division with \( x = 3 \):
\[
\begin{array}{r|rrrrr}
3 & 1 & 2 & -13 & -14 & 24 \\
& & 3 & 15 & 6 & -24 \\
\hline
& 1 & 5 & 2 & -8 & 0 \\
\end{array}
\]
Synthetic division with \( x = -4 \):
\[
\begin{array}{r|rrrr}
-4 & 1 & 5 & 2 & -8 \\
& & -4 & -4 & 8 \\
\hline
& 1 & 1 & -2 & 0 \\
\end{array}
\]
The depressed equation is:
\[
x^2 + x – 2 = 0
\]
Factoring the quadratic equation:
\[
x^2 + 2x – x – 2 = 0
\]
\[
x(x + 2) – 1(x + 2) = 0
\]
\[
(x + 2)(x – 1) = 0
\]
Therefore, the roots are:
\[
x = -2 \quad \text{or} \quad x = 1
\]
Thus, the roots of the given equation are:
\[
x = 3, \quad x = -4, \quad x = -2, \quad \text{and} \quad x = 1
\]
Next is Solutions of Exercise 2.7. We hope these detailed solutions have helped you understand the key concepts and techniques involved in Exercise 2.6 Class 10 Math. Synthetic division is a valuable method that simplifies polynomial calculations, allowing you to solve complex equations more easily. If you want to revisit previous concepts here is exercise 2.5 . If you have any remaining doubts or want additional explanations, be sure to check out the YouTube video lecture linked below. This video provides a step-by-step walkthrough of this exercise, offering further insights to strengthen your understanding. Keep practicing, and soon, synthetic division will become a straightforward tool in your math toolkit! Complete video lectures of unit 2 class 10 are here