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secondary 4 | A Maths
6 Answers Below
Anyone can contribute an answer, even non-tutors.
Thank you
Date Posted: 4 years ago
Views: 255
Seen. Later. The technique is to guess a number. Remember, fractional guesses are not in your syllabus.
Can teach me?
Will send one question first
I mean the answer to the first question first
Ok
Teach me the easy and easy to remember steps
Typically the numbers to guess are factors of the constant term, noth positive and negative.
So, if the cubic expression is ax3 + bx2 + cx + d, then the numbers to guess are factors of d, both positive and negative.
Not within your syllabus, but other guessable tries are factors of d/a or a/d and their negatives.
So, if the cubic expression is ax3 + bx2 + cx + d, then the numbers to guess are factors of d, both positive and negative.
Not within your syllabus, but other guessable tries are factors of d/a or a/d and their negatives.
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The relevant steps to solve cubic equations are as follows.
1. Guess a number which fits the equation. Typically whole numbers for your syllabus. Note that if the equation is not cubic, but say power 4, we need to guess two numbers. If power 5, three numbers. And so on.
2. Transform the guessed number into a factor. For example, if x = 1 is a solution, then (x - 1) is a factor.
3. Perform long division of the cubic expression with the factor to obtain a quotient in the form Ax2 + Bx + C and then factorise the expression. Alternatively, you can do comparing coefficients method. Since you have factored out a linear expression from a cubic equation, you will obtain a quadratic term (as per what I wrote).
4. We proceed to solve the equation in the factorised form.
1. Guess a number which fits the equation. Typically whole numbers for your syllabus. Note that if the equation is not cubic, but say power 4, we need to guess two numbers. If power 5, three numbers. And so on.
2. Transform the guessed number into a factor. For example, if x = 1 is a solution, then (x - 1) is a factor.
3. Perform long division of the cubic expression with the factor to obtain a quotient in the form Ax2 + Bx + C and then factorise the expression. Alternatively, you can do comparing coefficients method. Since you have factored out a linear expression from a cubic equation, you will obtain a quadratic term (as per what I wrote).
4. We proceed to solve the equation in the factorised form.
Date Posted:
4 years ago
Can help me out the rest?
Ok, one at a time, but not all today
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Same idea.
Date Posted:
4 years ago
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Q6c, Q6d
Similar techniques. The other tutor has done a good job for Q6e, so I will be skipping Q6e.
Similar techniques. The other tutor has done a good job for Q6e, so I will be skipping Q6e.
Date Posted:
4 years ago
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Q6f
Date Posted:
4 years ago
14.52, 0.48 (I never read the round off to 2 dp instruction)
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Q6g, Q6h
Date Posted:
4 years ago
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