PRACTICE QUESTIONS
ఈ అధ్యాయం విద్యార్థులకు వివిధ గణిత భావనలపై సమగ్ర అభ్యాసాన్ని అందిస్తుంది. బీజగణిత సమీకరణాలను కారకాలుగా విభజించడం, బహుపదుల గుణకాలను కనుగొనడం మరియు కారక సిద్ధాంతాన్ని ఉపయోగించి సమస్యలను పరిష్కరించడం వంటి అంశాలు ఇందులో ఉన్నాయి. ఈ ప్రాక్టీస్ ప్రశ్నలు విద్యార్థులు తమ సమస్య పరిష్కార నైపుణ్యాలను పెంపొందించుకోవడానికి, తార్కిక ఆలోచనను అభివృద్ధి చేసుకోవడానికి మరియు గణిత భావనలపై లోతైన అవగాహన పొందడానికి తోడ్పడతాయి. ఇది పరీక్షలలో మంచి మార్కులు సాధించడానికి కీలకమైన పునాదిని ఏర్పరుస్తుంది.
Factor Theorem and Remainder Theorem: Revision & Application
The Remainder Theorem and Factor Theorem are fundamental for polynomial factorisation. They help us find factors systematically.
- Remainder Theorem: If a polynomial \(P(x)\) is divided by a linear polynomial \((x - a)\), then the remainder is \(P(a)\).
- Application: Quickly find the remainder without performing long division.
- Factor Theorem: A special case of the Remainder Theorem.
- If \(P(a) = 0\), then \((x - a)\) is a factor of \(P(x)\).
- Conversely, if \((x - a)\) is a factor of \(P(x)\), then \(P(a) = 0\).
- Application: Crucial for finding linear factors of polynomials, especially cubic and quartic ones.
Steps to Apply Factor Theorem:
- Identify potential roots: For a polynomial \(P(x) = a_n x^n + ... + a_1 x + a_0\) with integer coefficients, any rational root \(p/q\) must have \(p\) as a factor of the constant term \(a_0\) and \(q\) as a factor of the leading coefficient \(a_n\).
- For monic polynomials (leading coefficient is 1), potential integer roots are factors of the constant term.
- Test potential roots: Substitute these values into \(P(x)\).
- Find a factor: If \(P(a) = 0\) for some value \(a\), then \((x - a)\) is a factor.
- Divide the polynomial: Use polynomial long division or synthetic division to divide \(P(x)\) by the found factor \((x - a)\). The quotient will be a polynomial of lower degree.
- Factorise the quotient: Continue factorising the quotient until all factors are found.
Important Identities for Factorisation:
- \((a+b)^2 = a^2 + 2ab + b^2\)
- \((a-b)^2 = a^2 - 2ab + b^2\)
- \(a^2 - b^2 = (a-b)(a+b)\)
- \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\)
- \((a-b)^3 = a^3 - b^3 - 3ab(a-b)\)
- \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\)
- \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)
- \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
- If \(a+b+c=0\), then \(a^3+b^3+c^3 = 3abc\). This is a very common exam question type!
When factorising cubic polynomials, always start by checking factors of the constant term (e.g., \(\pm 1, \pm 2, \pm 3, ...\)) using the Factor Theorem. This gives you the first linear factor.
Students often forget to check negative factors when applying the Factor Theorem. Remember to test both \(x=a\) and \(x=-a\).
Factorisation of Cubic Polynomials
Factorising cubic polynomials \(ax^3 + bx^2 + cx + d\) often involves a combination of the Factor Theorem and polynomial division.
Method 1: Using Factor Theorem and Long Division
- Find a root: Use the Factor Theorem to find one value \(a\) such that \(P(a) = 0\). This means \((x-a)\) is a factor.
- Divide: Divide \(P(x)\) by \((x-a)\) using polynomial long division. The quotient will be a quadratic polynomial.
- Factorise the quadratic: Factorise the resulting quadratic polynomial (e.g., by splitting the middle term, using the quadratic formula, or identities).
Method 2: Grouping (if applicable)
- Sometimes, a cubic polynomial can be factorised by grouping terms.
- Example: \(ax^3 + bx^2 + cx + d = x^2(ax+b) + (cx+d)\).
- This works if \((ax+b)\) and \((cx+d)\) share a common factor or can be made to share one.
Example Strategy for \(x^3 + 13x^2 + 32x + 20\):
- Possible factors of constant term (20): \(\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20\).
- Test \(x=-1\): \(P(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20 = -1 + 13 - 32 + 20 = 0\).
- So, \((x+1)\) is a factor.
- Divide \(x^3 + 13x^2 + 32x + 20\) by \((x+1)\):
- Using long division (or synthetic division):
\( (x^3 + 13x^2 + 32x + 20) \div (x+1) = x^2 + 12x + 20 \)
- Factorise the quadratic \(x^2 + 12x + 20\):
- Split the middle term: \(x^2 + 10x + 2x + 20 = x(x+10) + 2(x+10) = (x+2)(x+10)\).
- Final factorisation: \((x+1)(x+2)(x+10)\).
Special Case: Sum/Difference of Cubes
- If the polynomial is in the form \(a^3 \pm b^3\), use the identities directly.
- Example: \(8x^3 + 27 = (2x)^3 + 3^3 = (2x+3)((2x)^2 - (2x)(3) + 3^2) = (2x+3)(4x^2 - 6x + 9)\).
Polynomial long division is a key skill here. Practice it until you can do it quickly and accurately. Synthetic division is a faster alternative for dividing by linear factors \((x-a)\).
Factorisation of Quartic Polynomials
Factorising quartic polynomials \(ax^4 + bx^3 + cx^2 + dx + e\) typically involves finding two linear factors using the Factor Theorem, then dividing twice to reduce it to a quadratic.
Steps for Quartic Factorisation:
- Find first root: Use the Factor Theorem to find a value \(a\) such that \(P(a) = 0\). This gives the factor \((x-a)\).
- Divide: Divide \(P(x)\) by \((x-a)\). The quotient will be a cubic polynomial, say \(Q(x)\).
- Find second root: Apply the Factor Theorem again to \(Q(x)\) to find a value \(b\) such that \(Q(b) = 0\). This gives the factor \((x-b)\).
- Divide again: Divide \(Q(x)\) by \((x-b)\). The quotient will be a quadratic polynomial, say \(R(x)\).
- Factorise the quadratic: Factorise \(R(x)\) into two linear factors (if possible).
Example Strategy for \(x^4 + x^3 - 7x^2 - x + 6\):
- Possible factors of constant term (6): \(\pm 1, \pm 2, \pm 3, \pm 6\).
- Test \(x=1\): \(P(1) = 1^4 + 1^3 - 7(1)^2 - 1 + 6 = 1 + 1 - 7 - 1 + 6 = 0\).
- So, \((x-1)\) is a factor.
- Divide \(x^4 + x^3 - 7x^2 - x + 6\) by \((x-1)\):
- Quotient: \(x^3 + 2x^2 - 5x - 6\).
- Test \(x=-1\) on the quotient \(Q(x) = x^3 + 2x^2 - 5x - 6\):
- \(Q(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0\).
- So, \((x+1)\) is a factor of \(Q(x)\).
- Divide \(x^3 + 2x^2 - 5x - 6\) by \((x+1)\):
- Quotient: \(x^2 + x - 6\).
- Factorise the quadratic \(x^2 + x - 6\):
- \(x^2 + 3x - 2x - 6 = x(x+3) - 2(x+3) = (x-2)(x+3)\).
- Final factorisation: \((x-1)(x+1)(x-2)(x+3)\).
Quartic Factorisation by Substitution (for specific forms)
- For polynomials like \(ax^4 + bx^2 + c\) (biquadratic form), substitute \(y = x^2\).
- Example: \(x^4 + 5x^2 + 4\)
- Let \(y = x^2\). Then \(y^2 + 5y + 4\).
- Factorise: \(y^2 + 4y + y + 4 = y(y+4) + 1(y+4) = (y+1)(y+4)\).
- Substitute back \(y=x^2\): \((x^2+1)(x^2+4)\).
- Note: These factors \((x^2+1)\) and \((x^2+4)\) cannot be factorised further into real linear factors. If the question asks for real factors, this is the final answer. If complex factors are allowed (not in Class 9), then \((x-i)(x+i)(x-2i)(x+2i)\).
For quartic polynomials, finding the first two factors is usually the most challenging part. Once you get to a quadratic, factorisation is straightforward.
Don't forget to check if the quadratic factor obtained after division can be factorised further. Many students stop after finding the first linear factor.
Advanced Factorisation Techniques & Special Cases
Beyond the standard Factor Theorem and grouping, some problems require specific techniques or recognition of patterns.
Factorisation by Grouping (Advanced)
- Sometimes, terms need to be rearranged or split to facilitate grouping.
- Example: \(9z^3 - 27z^2 - 100z + 300\)
- Group: \(9z^2(z-3) - 100(z-3)\)
- Common factor \((z-3)\): \((z-3)(9z^2 - 100)\)
- Apply \(a^2 - b^2\) identity: \((z-3)((3z)^2 - 10^2) = (z-3)(3z-10)(3z+10)\).
Factorisation of Polynomials with Irrational Coefficients (Rare but possible)
- Example: \(x^3 - x^2 - (2 - \sqrt{2})x + \sqrt{2}\)
- Try \(x=1\): \(P(1) = 1 - 1 - (2 - \sqrt{2}) + \sqrt{2} = -2 + \sqrt{2} + \sqrt{2} = -2 + 2\sqrt{2} \ne 0\)
- Try \(x=-1\): \(P(-1) = -1 - 1 - (2 - \sqrt{2})(-1) + \sqrt{2} = -2 + (2 - \sqrt{2}) + \sqrt{2} = -2 + 2 - \sqrt{2} + \sqrt{2} = 0\)
- So, \((x+1)\) is a factor.
- Divide \(x^3 - x^2 - (2 - \sqrt{2})x + \sqrt{2}\) by \((x+1)\).
- Quotient: \(x^2 - 2x + \sqrt{2}\).
- This quadratic might not factorise easily with splitting the middle term. It might require the quadratic formula if further factorisation is needed (not typical for Class 9).
- In the given question, just determining if \((x+1)\) is a factor is sufficient.
Factorisation by Splitting Terms (when Factor Theorem is hard to apply directly)
- Sometimes, you might need to rewrite terms to create common factors.
- Example: \(2y^3 + y^2 - 2y - 1\)
- Group: \(y^2(2y+1) - 1(2y+1)\)
- Common factor \((2y+1)\): \((2y+1)(y^2-1)\)
- Apply \(a^2 - b^2\) identity: \((2y+1)(y-1)(y+1)\).
General Strategy for Factorisation Problems:
- Look for common factors: Always the first step.
- Count terms:
- Two terms: Check for \(a^2-b^2\), \(a^3 \pm b^3\).
- Three terms (quadratic): Split the middle term, or use quadratic formula.
- Four terms: Try grouping. If not, use Factor Theorem to find a linear factor, then divide.
- More than four terms: Use Factor Theorem to reduce the degree, then repeat.
- Use identities: Recognise patterns like perfect squares, sum/difference of cubes.
- Factor Theorem: For polynomials of degree 3 or higher, this is your go-to method.
- Polynomial Division: Essential after finding a factor using the Factor Theorem.
Always check if the constant term has many factors. This indicates many possible values to test for the Factor Theorem. Start with \(\pm 1, \pm 2\).
If a polynomial has coefficients that are large, or if the constant term has many factors, sometimes the roots are fractions (e.g., \(p/q\)). Remember to test these too, especially if the leading coefficient is not 1.
