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How do factor a trinomial expression with a leading coefficient not equal to 1?
To factor a trinomial in the form ax2 + bx + c, where a does not equal 1, the easiest process is called "factoring by grouping". To factor by grouping, you must change the trinomial into an equivalent tetranomial by rewriting the middle term (bx) as the sum of two terms. There is a specific way to do this, as demonstrated in the example.Take the quadratic trinomial 5x2 + 11x + 21. Find the product of a and c, or 5*2 = 10.2. Find factors of ac that when added together give you b, in this case 10 and 1.3. Rewrite the middle term as the sum of the two factors (5x2 + 10x + x + 2).4. Group terms with common factors and factor these groups.5x2 + x + 10x + 2x(5x + 1) + 2(5x + 1)5. Factor the binomial in the parentheses out of the whole polynomial, leaving you with the product of two binomials. 5x2 + 11x + 2 = (x + 2)(5x + 1)Notes:1. The same process is done if there are any minus signs in the trinomial, just be careful when factoring out a negative from a positive or vice versa.2. If you have a tetranomial on its own, you can skip the rewriting process and just factor the whole polynomial by grouping from the start.3. As in factoring any polynomial, always factor out the GCF first, then factor the remaining polynomial if necessary.4. Always look for patterns, like the difference of squares or square of a binomial, while factoring. It will save a lot of time.
Example of square of a trinomial?
If you want to know how to square a trinomial, you should first know the basic. (a+b+c)^2=? you have to square the first three terms then multiply 2 to the last three terms. All you have to do is to remember that a square of trinomial has 6 terms in the answer
How can a trinomial be recognized as a perfect square?
You can easily identify it.The first and last term are perfect squares.Example: X2 + 2xy + y2The first and last term are Positive.* * * * *That is rubbish.The first and last terms of x2 + x + 1 are perfect squares but the trinomial is not. In fact, it has no real factors.If the trinomial is written in the form ax2 + bx + c , then it is a perfect square if b2 = 4ac
What is the definition of quadratic equation of factoring?
its easy first,xczxczxczxczxc....ERROR..vxbdxv
What strategies is appropriate for factoring polynomials with 4 terms?
A strategy that would be appropriate in factoring polynomials with 4 terms would be by grouping where you first determine if the polynomial can be factored by a group.
When you are factoring a trinomial with a leading coefficient other than 1 what is the first step?
find all the factors of the constant term
In factoring a trinomial with a leading coefficient other than 1 the first step is to look for a factor in each term?
Common Apex
How do you complete the square to make a binomial a perfect-square trinomial?
The binomial usually has an x2 term and an x term, so we complete the square by adding a constant term. If the coefficient of x2 is not 1, we divide the binomial by that coefficient first (we can multiply the trinomial by it later). Then we divide the coefficient of x by 2 and square that. That is the constant that we need to add to get the perfect square trinomial. Then just multiply that trinomial by the original coefficient of x2.
In which situation would you most likely factor out -1 from a trinomial?
If the coefficient of the highest power of a variable of interest is negative.
What is the factor of the trinomial 3m2 plus 11mn plus 6n2?
To find the factors of the trinomial (3m^2 + 11mn + 6n^2), we need to break it down into two binomials. First, we find two numbers that multiply to the product of the leading coefficient and constant term, which are (3 \times 6 = 18). Then, we look for two numbers that add up to the middle coefficient, which is 11. The factors are ((3m + 2n)(m + 3n)).
What is the first thing you should look for when factoring?
Is the coefficient of the square a prime number? eg if the equation begins 3a2 then the factors must be (3a +/- x)(a +/- y)
How do factor a trinomial expression with a leading coefficient not equal to 1?
To factor a trinomial in the form ax2 + bx + c, where a does not equal 1, the easiest process is called "factoring by grouping". To factor by grouping, you must change the trinomial into an equivalent tetranomial by rewriting the middle term (bx) as the sum of two terms. There is a specific way to do this, as demonstrated in the example.Take the quadratic trinomial 5x2 + 11x + 21. Find the product of a and c, or 5*2 = 10.2. Find factors of ac that when added together give you b, in this case 10 and 1.3. Rewrite the middle term as the sum of the two factors (5x2 + 10x + x + 2).4. Group terms with common factors and factor these groups.5x2 + x + 10x + 2x(5x + 1) + 2(5x + 1)5. Factor the binomial in the parentheses out of the whole polynomial, leaving you with the product of two binomials. 5x2 + 11x + 2 = (x + 2)(5x + 1)Notes:1. The same process is done if there are any minus signs in the trinomial, just be careful when factoring out a negative from a positive or vice versa.2. If you have a tetranomial on its own, you can skip the rewriting process and just factor the whole polynomial by grouping from the start.3. As in factoring any polynomial, always factor out the GCF first, then factor the remaining polynomial if necessary.4. Always look for patterns, like the difference of squares or square of a binomial, while factoring. It will save a lot of time.
How do you factor polynomials?
A few examples..** First, taking out a common factor:1:x3 + 3xx(x2 + 3) -This is factored because you have taken an "x" away from both terms... to put it back into its original form just multiply the "x" by x2 and 3.2:6x2 + 16x + 82(3x2 + 8x + 4) -All of the numbers were divisible by 2 so you can take 2 out of all of the terms.** Factoring a Difference of Squares:x2 - 16 = (x)2 - 42 = (x - 4)(x + 4).** An 'easy' trinomial (the coefficient of x is 1):x2 + 5x + 6 = (x + 3)(x + 2). -Notice that 2+3=5 and (2)(3) =6.** A 'hard' trinomial (the coefficient of x2 is not 1):8x2 - 2x - 21 = (2x + 3)(4x - 7). -There are several ways of organizing a trial and error process to factor a case like this.** Factoring by grouping (Check for this when there are 4 terms.)6x3 - 2x2 + 15x - 5 = 2x2(3x - 1) + 5(3x - 1) = (2x2 + 5)(3x - 1).
What are the zeros of the quadratic equation whose leading coefficient is -1 and coefficient of x is -7 and constant term is -12?
In other words, the zeroes of -x2 - 7x - 12.First, multiply by -1: x2 + 7x + 12.The new leading coefficient is 1, so the factors take the form (x + _)(x + _), where the two blanked-out numbers add up to 7 and multiply to 12.It's easier to try factoring 12 and adding the factors:1 + 12 = 132 + 6 = 83 + 4 = 7That last one shows us that the factors are (x + 3)(x + 4), and the zeroes are -3 and -4.
Example of square of a trinomial?
If you want to know how to square a trinomial, you should first know the basic. (a+b+c)^2=? you have to square the first three terms then multiply 2 to the last three terms. All you have to do is to remember that a square of trinomial has 6 terms in the answer
How do you factor q2-7q plus 12?
1. When factoring first always look for a GCF (greatest common factor). If each term has a greatest common factor, factor it out in from using parenthesis first. This problem does not have a GCF. 2. Next, since this is a trinomial, many times we can factor it down using backwards FOIL (First, Outter, Inner, Last). 3. To do this always put down two sets of parenthesis. (we do this because we are looking to factor into two binomials) ( )( ) 4. Next we complete the fist term in each set of parenthesis. The first term is simply going to be the variable we are using in the problem. In this problem the variable is q. (q )(q ) 5. Then find the factors of the last term (+12) in which the sum is equal to the coefficient of the middle term (-7). These factors are -3 and -4. 6. Complete the factoring by putting these factors into the second part of the parenthesis. (q - 3)(q - 4) * If you want to make sure you are correct, multiply you answer out and see if you get the same trinomial you started with.
Can all quadratic equations be solved?
Well, that depends on what you mean "solve by factoring." For any quadratic equation, it is possible to factor the quadratic, and then the roots can be recovered from the factors. So in the very weak sense that every quadratic can be solved by a method that involves getting the factors and recovering the roots from them, all quadratic equations can be solved by factoring. However, in most cases, the only way of factoring the quadratic in the first place is to first find out what its roots are, and then use the roots to factor the quadratic (any quadratic polynomial can be factored as k(x - r)(x - s), where k is the leading coefficient of the polynomial and r and s are its two roots), in which case trying to recover the roots from the factors is redundant (since you had to know what the roots were to get the factors in the first place). So to really count as solving by factoring, it makes sense to require that the solution method obtains the factors by means that _don't_ require already knowing the roots of the polynomial. And in this sense, most quadratic equations are not solvable through factoring.
