1. Determine which of the following polynomials has (x + 1) a factor:
(i) \(x^3+x^2+x+1\)
Solution:
Let \(p(x) = x^3+x^2+x+1\)
The zero of x+1 is -1. [x+1 = 0 means x = -1]
\(p(−1) = (−1)^3+(−1)^2+(−1)+1\)
\(= −1+1−1+1\)
\(= 0\)
∴ By factor theorem, x+1 is a factor of \(x^3+x^2+x+1\)
(ii) \(x^4+x^3+x^2+x+1\)
Solution:
Let \(p(x)= x^4+x^3+x^2+x+1\)
The zero of x+1 is -1. [x+1= 0 means x = -1]
\(p(−1) = (−1)^4+(−1)^3+(−1)^2+(−1)+1\)
\(= 1−1+1−1+1\)
\(= 1 ≠ 0\)
∴ By factor theorem, x+1 is not a factor of \(x^4+x^3+x^2+x+1\)
(iii) \(x^4+3x^3+3x^2+x+1\)
Solution:
Let \(p(x)= x^4+3x^3+3x^2+x+1\)
The zero of x+1 is -1.
\(p(−1)=(−1)^4+3(−1)^3+3(−1)^2+(−1)+1\)
\(=1−3+3−1+1\)
\(=1 ≠ 0\)
∴ By factor theorem, x+1 is not a factor of \(x^4+3x^3+3x^2+x+1\)
(iv) \(x^3 – x^2– (2+√2)x +√2\)
Solution:
Let \(p(x) = x^3–x^2–(2+√2)x +√2\)
The zero of x+1 is -1.
\(p(−1) = (-1)^3–(-1)^2–(2+√2)(-1) + √2 = −1−1+2+√2+√2\)
\(= 2√2 ≠ 0\)
∴ By factor theorem, x+1 is not a factor of \(x^3–x^2–(2+√2)x +√2\)
2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) \(p(x) = 2x^3+x^2–2x–1, g(x) = x+1\)
Solution:
\(p(x) = 2x^3+x^2–2x–1, g(x) = x+1\)
\(g(x) = 0\)
\( x+1 = 0\)
\( x = −1\)
∴ Zero of g(x) is -1.
\(p(−1) = 2(−1)^3+(−1)^2–2(−1)–1\)
= −2+1+2−1
= 0
∴ By factor theorem, g(x) is a factor of p(x).
(ii) \(p(x)=x^3+3x^2+3x+1, g(x) = x+2\)
Solution:
\(p(x) = x^3+3x^2+3x+1, g(x) = x+2\)
g(x) = 0
\( x+2 = 0\)
\( x = −2\)
∴ Zero of g(x) is -2.
\(p(−2) = (−2)^3+3(−2)^2+3(−2)+1\)
\(= −8+12−6+1\)
\(= −1 ≠ 0\)
∴ By factor theorem, g(x) is not a factor of p(x).
(iii) \(p(x)=x^3–4x^2+x+6, g(x) = x–3\)
Solution:
\(p(x) = x^3–4x^2+x+6, g(x) = x -3\)
g(x) = 0
\( x−3 = 0\)
\( x = 3\)
∴ Zero of g(x) is 3.
\(p(3) = (3)^3−4(3)^2+(3)+6\)
\(= 27−36+3+6\)
\(= 0\)
∴ By factor theorem, g(x) is a factor of p(x).
3. Find the value of k, if x–1 is a factor of p(x) in each of the following cases:
(i) \(p(x) = x^2+x+k\)
Solution:
If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem
\( (1)^2+(1)+k = 0\)
\(1+1+k = 0\)
\( 2+k = 0\)
\( k = −2\)
(ii) p(x) = 2x2+kx+√2
Solution:
If x-1 is a factor of p(x), then p(1) = 0
\( 2(1)^2+k(1)+√2 = 0\)
\( 2+k+√2 = 0\)
\( k = −(2+\sqrt2)\)
(iii) \(p(x) = kx^2–\sqrt2x+1\)
Solution:
If x-1 is a factor of p(x), then p(1)=0
By Factor Theorem
\( k(1)^2-\sqrt{2} \times 1+1=0\)
\(k = \sqrt{2}-1\)
(iv) \(p(x)=kx^2–3x+k\)
Solution:
If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem
\( k(1)^2–3(1)+k = 0\)
\( k−3+k = 0\)
\( 2k−3 = 0\)
\( k= {3\over 2}\)
4. Factorise:
(i) \(12x^2–7x+1\)
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = -7 and product =1×12 = 12
We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 12]
\(12x^2–7x+1= 12x^2-4x-3x+1\)
\( 4x(3x-1)-1(3x-1)\)
\( (4x-1)(3x-1)\)
(ii) \(2x^2+7x+3\)
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = 7 and product = 2×3 = 6
We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6]
\(2x^2+7x+3 = 2x^2+6x+1x+3\)
\(= 2x (x+3)+1(x+3)\)
\(= (2x+1)(x+3)\)
(iii) \(6x^2+5x-6\)
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = 5 and product = 6×-6 = -36
We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36]
\(6x^2+5x-6 = 6x^2+9x–4x–6\)
\(= 3x(2x+3)–2(2x+3)\)
\(= (2x+3)(3x–2)\)
(iv) \(3x^2–x–4\)
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = -1 and product = 3×-4 = -12
We get -4 and 3 as the numbers [-4+3 = -1 and -4×3 = -12]
\(3x^2–x–4 = 3x^2–4x+3x–4\)
\(= x(3x–4)+1(3x–4)\)
\(= (3x–4)(x+1)\)
5. Factorise:
(i) \(x^3–2x^2–x+2\)
Solution:
Let \(p(x) = x^3–2x^2–x+2\)
Factors of 2 are ±1 and ± 2
\(p(x) = x^3–2x^2–x+2\)
\(p(−1) = (−1)3–2(−1)2–(−1)+2\)
\(= −1−2+1+2\)
\(= 0\)
Therefore, (x+1) is the factor of p(x)
\((x+1)(x^2–3x+2) = (x+1)(x^2–x–2x+2)\)
\(= (x+1)(x(x−1)−2(x−1))\)
\(= (x+1)(x−1)(x-2)\)
(ii)\( x^3–3x^2–9x–5\)
Solution:
Let \(p(x) = x^3–3x^2–9x–5\)
Factors of 5 are ±1 and ±5
By the trial method, we find that
p(5) = 0
So, (x-5) is factor of p(x)
\(p(x) = x^3–3x^2–9x–5\)
\(p(5) = (5)^3–3(5)^2–9(5)–5\)
\(= 125−75−45−5\)
\(= 0\)
Therefore, \((x-5)\) is the factor of p(x)
\((x−5)(x^2+2x+1) = (x−5)(x^2+x+x+1)\)
\(= (x−5)(x(x+1)+1(x+1))\)
\(= (x−5)(x+1)(x+1)\)
(iii) \(x^3+13x^2+32x+20\)
Solution:
Let \(p(x) = x^3+13x^2+32x+20\)
Factors of 20 are \(±1, ±2, ±4, ±5, ±10\) and \(±20\)
By the trial method, we find that
\(p(-1) = 0\)
So, \((x+1)\) is factor of p(x)
\(p(x)= x^3+13x^2+32x+20\)
\(p(-1) = (−1)^3+13(−1)^2+32(−1)+20\)
\(= −1+13−32+20\)
\(= 0\)
Therefore, \((x+1)\) is the factor of p(x)
\((x+1)(x^2+12x+20) = (x+1)(x^2+2x+10x+20)\)
\(= (x+1)x(x+2)+10(x+2)\)
\(= (x+1)(x+2)(x+10)\)
(iv) \(2y^3+y^2–2y–1\)
Solution:
Let \(p(y) = 2y^3+y^2–2y–1\)
Factors \(= 2×(−1)= -2\) are \(±1\) and \(±2\)
By the trial method, we find that
\(p(y) = 2y^3+y^2–2y–1\)
\(p(1) = 2(1)^3+(1)^2–2(1)–1 = 2+1−2 = 0\)
Therefore, (y-1) is the factor of p(y)
\((y−1)(2y^2+3y+1) = (y−1)(2y^2+2y+y+1)\)
\(= (y−1)(2y(y+1)+1(y+1))\)
\(= (y−1)(2y+1)(y+1)\)
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