Mathematics 11 (Punjab Board)
Exercise 9.1 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 9: Division of Polynomials. This exercise covers long division, the remainder theorem, the factor theorem, and synthetic division.
Question 1: Find the remainder and the quotient by simplifying the following: (i) \((3x^2-x+2) \div (x-1)\)
Long division: sab se pehle \(3x^2 \div x = 3x\), phir multiply karke subtract karenge. Aakhir mein check karenge ke \((x-1)\times\) quotient + remainder = dividend.
Step 1: \(3x^2 \div x = 3x\). \(3x(x-1) = 3x^2-3x\). Subtract: \((3x^2-x+2)-(3x^2-3x) = 2x+2\).
Step 2: \(2x \div x = 2\). \(2(x-1) = 2x-2\). Subtract: \((2x+2)-(2x-2) = 4\).
\[
3x^2-x+2 = (x-1)(3x+2) + 4
\]
Answer: Quotient \(= 3x+2\), remainder \(= 4\).
Question 1(ii): Find the quotient and remainder: \((x^3+12x^2-3x+4) \div (x-2)\)
Teen steps honge kyunke dividend degree 3 ka hai. Har step mein leading term divide karke subtract karenge.
Step 1: \(x^3 \div x = x^2\). \(x^2(x-2) = x^3-2x^2\). Subtract: \(14x^2-3x+4\).
Step 2: \(14x^2 \div x = 14x\). \(14x(x-2) = 14x^2-28x\). Subtract: \(25x+4\).
Step 3: \(25x \div x = 25\). \(25(x-2) = 25x-50\). Subtract: \(54\).
\[
x^3+12x^2-3x+4 = (x-2)(x^2+14x+25) + 54
\]
Answer: Quotient \(= x^2+14x+25\), remainder \(= 54\).
Question 1(iii): Find the quotient and remainder: \((x^4-5x^3-8x^2+13x+12) \div (x+6)\)
Note: the printed polynomial does not match the book's own answer key (which fits \(x^4+7x^3+4x^2-11x+24\)). The long division below is for the polynomial as printed, verified by back-multiplication.
Step 1: \(x^4 \div x = x^3\). \(x^3(x+6) = x^4+6x^3\). Subtract: \(-11x^3-8x^2+13x+12\).
Step 2: \(-11x^3 \div x = -11x^2\). \(-11x^2(x+6) = -11x^3-66x^2\). Subtract: \(58x^2+13x+12\).
Step 3: \(58x^2 \div x = 58x\). \(58x(x+6) = 58x^2+348x\). Subtract: \(-335x+12\).
Step 4: \(-335x \div x = -335\). \(-335(x+6) = -335x-2010\). Subtract: \(2022\).
\[
x^4-5x^3-8x^2+13x+12 = (x+6)(x^3-11x^2+58x-335) + 2022
\]
Check: \(f(-6) = 1296+1080-288-78+12 = 2022\) ✓ (remainder theorem).
Answer: Quotient \(= x^3-11x^2+58x-335\), remainder \(= 2022\).
Question 1(iv): Find the quotient and remainder: \((5x^4-3x^3+2x^2-1) \div (x^2+4)\)
Divisor degree 2 ka hai, to remainder ki degree 1 se kam hogi (linear ya constant). Missing \(x\) term ke liye \(0x\) likh kar division karenge.
Step 1: \(5x^4 \div x^2 = 5x^2\). \(5x^2(x^2+4) = 5x^4+20x^2\). Subtract: \(-3x^3-18x^2+0x-1\).
Step 2: \(-3x^3 \div x^2 = -3x\). \(-3x(x^2+4) = -3x^3-12x\). Subtract: \(-18x^2+12x-1\).
Step 3: \(-18x^2 \div x^2 = -18\). \(-18(x^2+4) = -18x^2-72\). Subtract: \(12x+71\).
\[
5x^4-3x^3+2x^2-1 = (x^2+4)(5x^2-3x-18) + (12x+71)
\]
Answer: Quotient \(= 5x^2-3x-18\), remainder \(= 12x+71\).
Question 1(v): Find the quotient and remainder: \((3x^4-5x^3+4x-6) \div (x^2-3x+5)\)
Missing \(x^2\) term ke liye \(0x^2\) likhenge. Teen steps ke baad remainder \(-25x+9\) bachega (degree 1, jo divisor ki degree 2 se kam hai — theek hai).
Step 1: \(3x^4 \div x^2 = 3x^2\). \(3x^2(x^2-3x+5) = 3x^4-9x^3+15x^2\). Subtract: \(4x^3-15x^2+4x-6\).
Step 2: \(4x^3 \div x^2 = 4x\). \(4x(x^2-3x+5) = 4x^3-12x^2+20x\). Subtract: \(-3x^2-16x-6\).
Step 3: \(-3x^2 \div x^2 = -3\). \(-3(x^2-3x+5) = -3x^2+9x-15\). Subtract: \(-25x+9\).
\[
3x^4-5x^3+4x-6 = (x^2-3x+5)(3x^2+4x-3) + (-25x+9)
\]
Answer: Quotient \(= 3x^2+4x-3\), remainder \(= -25x+9\).
Question 2: Use the remainder theorem to find the remainder when the first polynomial is divided by the second: (i) \(x^2+5x+6\) by \(x-2\)
Remainder theorem: \(f(x) \div (x-a)\) ka remainder \(f(a)\) hai. Yahan \(a = 2\), to \(f(2)\) nikalenge.
\[
f(2) = (2)^2 + 5(2) + 6 = 4 + 10 + 6 = 20
\]
Answer: Remainder \(= 20\).
Question 2(ii): Find the remainder when \(x^3+5x^2+6\) is divided by \(x+1\).
\(x+1 = x-(-1)\), to \(a = -1\). \(f(-1)\) nikalenge; even/odd powers ka khayal rakhen.
\[
f(-1) = (-1)^3 + 5(-1)^2 + 6 = -1 + 5 + 6 = 10
\]
Answer: Remainder \(= 10\).
Question 2(iii): Find the remainder when \(x^4+x^3+x^2+x+1\) is divided by \(x-1\).
\(a = 1\) rakh kar evaluate karenge — sab terms 1 ban jayenge.
\[
f(1) = 1 + 1 + 1 + 1 + 1 = 5
\]
Answer: Remainder \(= 5\).
Question 2(iv): Find the remainder when \(x^4+x^2+1\) is divided by \(x+3\).
\(a = -3\). \((-3)^4 = 81\) aur \((-3)^2 = 9\) — negative ki even power positive hoti hai.
\[
f(-3) = (-3)^4 + (-3)^2 + 1 = 81 + 9 + 1 = 91
\]
Answer: Remainder \(= 91\).
Question 2(v): Find the remainder when \(x^4+x^3+2\) is divided by \(x+2\).
\(a = -2\): \((-2)^4 = 16\), \((-2)^3 = -8\).
\[
f(-2) = (-2)^4 + (-2)^3 + 2 = 16 - 8 + 2 = 10
\]
Answer: Remainder \(= 10\).
Question 3: Use the factor theorem to determine whether the first polynomial is a factor of the second: (i) \(x+1\) and \(x^2-1\)
Factor theorem: \(x-a\) factor hai agar \(f(a) = 0\). \(x+1 = x-(-1)\), to \(f(-1)\) check karenge.
\[
f(-1) = (-1)^2 - 1 = 0
\]
Since the remainder is 0, \(x+1\) is a factor of \(x^2-1\).
Answer: Yes, \(x+1\) is a factor of \(x^2-1\).
Question 3(ii): Determine whether \(x-2\) is a factor of \(x^2-5x+6\).
\(f(2)\) nikalenge. Agar zero aaye to factor hai.
\[
f(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0
\]
Answer: Yes, \(x-2\) is a factor of \(x^2-5x+6\).
Question 3(iii): Determine whether \(x+1\) is a factor of \(x^3+x^2+x-3\).
\(f(-1)\) nikalenge.
\[
f(-1) = (-1)^3 + (-1)^2 + (-1) - 3 = -1 + 1 - 1 - 3 = -4 \ne 0
\]
Answer: No, \(x+1\) is not a factor of \(x^3+x^2+x-3\) (remainder \(-4\)).
Question 3(iv): Determine whether \(x-2\) is a factor of \(x^3+x^2-7x+2\).
\(f(2)\) nikalenge.
\[
f(2) = (2)^3 + (2)^2 - 7(2) + 2 = 8 + 4 - 14 + 2 = 0
\]
Answer: Yes, \(x-2\) is a factor of \(x^3+x^2-7x+2\).
Question 3(v): Determine whether \(x-3\) is a factor of \(x^4-3x^3+x^2-x+1\).
\(f(3)\) nikalenge. Note karen ke \(-3(3)^3 = -81\) aur \((3)^4 = 81\) cancel ho jayenge.
\[
f(3) = (3)^4 - 3(3)^3 + (3)^2 - 3 + 1 = 81 - 81 + 9 - 3 + 1 = 7 \ne 0
\]
Answer: No, \(x-3\) is not a factor of \(x^4-3x^3+x^2-x+1\) (remainder \(7\)).
Question 4: Use synthetic division to show that the given value is a zero of the polynomial, and use the result to factorize the polynomial completely: (i) \(x^3-7x+6\), \(x = 2\)
Note: the book prints \(x^2-7x+6\), but \(x = 2\) is not its zero (\(f(2) = -4\)). The intended polynomial is \(x^3-7x+6\), since \(f(2) = 8-14+6 = 0\). Missing \(x^2\) ke liye coefficient 0 likhenge.
\[
\begin{array}{r|rrrr}
2 & 1 & 0 & -7 & 6 \\
& & 2 & 4 & -6 \\ \hline
& 1 & 2 & -3 & 0
\end{array}
\]
Remainder 0, so \(x = 2\) is a zero. Quotient: \(x^2+2x-3 = (x+3)(x-1)\).
\[
x^3-7x+6 = (x-2)(x+3)(x-1)
\]
Answer: \((x-2)(x-1)(x+3)\).
Question 4(ii): Show that \(x = -4\) is a zero of \(x^3-28x-48\) and factorize completely.
Synthetic division \(-4\) se karenge. Quotient \(x^2-4x-12\) ko mazeed factorize karenge.
\[
\begin{array}{r|rrrr}
-4 & 1 & 0 & -28 & -48 \\
& & -4 & 16 & 48 \\ \hline
& 1 & -4 & -12 & 0
\end{array}
\]
Remainder 0, so \(x = -4\) is a zero. Quotient: \(x^2-4x-12 = (x-6)(x+2)\).
\[
x^3-28x-48 = (x+4)(x-6)(x+2)
\]
Answer: \((x+4)(x-6)(x+2)\).
Question 4(iii): Show that \(x = 2\) and \(x = -3\) are zeros of \(2x^4+7x^3-4x^2-27x-18\) and factorize completely.
Do martaba synthetic division: pehle 2 se, phir bache hue quotient ko \(-3\) se. Aakhir mein quadratic \(2x^2+5x+3\) ko factorize karenge.
\[
\begin{array}{r|rrrrr}
2 & 2 & 7 & -4 & -27 & -18 \\
& & 4 & 22 & 36 & 18 \\ \hline
& 2 & 11 & 18 & 9 & 0
\end{array}
\qquad
\begin{array}{r|rrrr}
-3 & 2 & 11 & 18 & 9 \\
& & -6 & -15 & -9 \\ \hline
& 2 & 5 & 3 & 0
\end{array}
\]
Both remainders are 0. Final quotient: \(2x^2+5x+3 = (2x+3)(x+1)\).
\[
2x^4+7x^3-4x^2-27x-18 = (x-2)(x+3)(2x+3)(x+1)
\]
Answer: \((x-2)(x+3)(2x+3)(x+1)\).
Question 5: Use synthetic division to find the quotient and the remainder when \(x^4-10x^2-2x+4\) is divided by \(x+3\).
\(x+3 = x-(-3)\), to \(-3\) se synthetic division karenge. Missing \(x^3\) ke liye coefficient 0 likhenge: coefficients \(1, 0, -10, -2, 4\).
\[
\begin{array}{r|rrrrr}
-3 & 1 & 0 & -10 & -2 & 4 \\
& & -3 & 9 & 3 & -3 \\ \hline
& 1 & -3 & -1 & 1 & 1
\end{array}
\]
Bottom row: quotient coefficients \(1, -3, -1, 1\) and remainder \(1\). Check: \(f(-3) = 81-90+6+4 = 1\) ✓.
Answer: Quotient \(= x^3-3x^2-x+1\), remainder \(= 1\).
Question 6: If \(x+1\) and \(x-2\) are factors of \(x^3-px^2+qx+2\), find the values of \(p\) and \(q\) using synthetic division.
Pehle factor theorem se do equations banayenge: \(f(-1) = 0\) aur \(f(2) = 0\). Phir solve karke \(p, q\) nikalenge. (Synthetic division se bhi verify ho sakta hai ke remainders zero hain.)
\[
f(-1) = -1-p-q+2 = 0 \Rightarrow p+q = 1 \quad \text{(i)}
\]
\[
f(2) = 8-4p+2q+2 = 0 \Rightarrow 4p-2q = 10 \Rightarrow 2p-q = 5 \quad \text{(ii)}
\]
Adding (i) and (ii): \(3p = 6 \Rightarrow p = 2\). From (i): \(q = 1-2 = -1\).
Verification by synthetic division with \(p = 2, q = -1\) (coefficients \(1, -2, -1, 2\)):
\[
\begin{array}{r|rrrr}
-1 & 1 & -2 & -1 & 2 \\
& & -1 & 3 & -2 \\ \hline
& 1 & -3 & 2 & 0
\end{array}
\qquad
\begin{array}{r|rrrr}
2 & 1 & -2 & -1 & 2 \\
& & 2 & 0 & -2 \\ \hline
& 1 & 0 & -1 & 0
\end{array}
\]
Both remainders are 0, confirming the factors.
Answer: \(p = 2\), \(q = -1\).
Question 7: When the polynomial \(4x^4+2x^3+kx^2+13\) is divided by \(x+1\), the remainder is 16. Find the value of \(k\).
Remainder theorem: \(f(-1) = 16\). Is mein \(k\) unknown hai; equation solve karenge.
\[
f(-1) = 4(-1)^4 + 2(-1)^3 + k(-1)^2 + 13 = 4 - 2 + k + 13 = 15 + k
\]
\[
15 + k = 16 \Rightarrow k = 1
\]
Answer: \(k = 1\).
Question 8: When the polynomial \(x^3+x^2+x+k\) is divided by \(x+1\), the remainder is 7. Find the value of \(k\).
\(f(-1) = 7\) rakhenge aur solve karenge.
\[
f(-1) = (-1)^3 + (-1)^2 + (-1) + k = -1 + 1 - 1 + k = k - 1
\]
\[
k - 1 = 7 \Rightarrow k = 8
\]
Answer: \(k = 8\).
Question 9: Use the factor theorem to find the values of \(p\) and \(q\) if \(x+1\) and \(x-2\) are factors of \(x^3+px^2+qx+3\).
\(f(-1) = 0\) aur \(f(2) = 0\) se do equations banengi. Yahan constants fractions mein aayenge — ghabrayen nahi, steps wese hi hain.
\[
f(-1) = -1+p-q+3 = 0 \Rightarrow p-q = -2 \quad \text{(i)}
\]
\[
f(2) = 8+4p+2q+3 = 0 \Rightarrow 4p+2q = -11 \Rightarrow 2p+q = -\frac{11}{2} \quad \text{(ii)}
\]
Adding (i) and (ii): \(3p = -\frac{15}{2} \Rightarrow p = -\frac{5}{2}\). From (i): \(q = p+2 = -\frac{1}{2}\).
Check: \(f(-1) = -1-\frac{5}{2}+\frac{1}{2}+3 = 0\) ✓; \(f(2) = 8-10-1+3 = 0\) ✓.
Answer: \(p = -\frac{5}{2}\), \(q = -\frac{1}{2}\).
Question 10: Use the factor theorem to find the values of \(a\) and \(b\) if \(-2\) and \(2\) are the roots of \(2x^3+4x^2+ax+b\).
Root ka matlab \(f(-2) = 0\) aur \(f(2) = 0\). Do equations se \(a, b\) nikalenge.
\[
f(-2) = 2(-8)+4(4)-2a+b = -16+16-2a+b = 0 \Rightarrow -2a+b = 0 \quad \text{(i)}
\]
\[
f(2) = 2(8)+4(4)+2a+b = 16+16+2a+b = 0 \Rightarrow 2a+b = -32 \quad \text{(ii)}
\]
Adding (i) and (ii): \(2b = -32 \Rightarrow b = -16\). From (i): \(-2a-16 = 0 \Rightarrow a = -8\).
Check: \(2x^3+4x^2-8x-16 = 2(x+2)(x-2)(x+2)\); roots \(-2, 2\) ✓.
Answer: \(a = -8\), \(b = -16\).
