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What is the difficulty in factoring a trinomial with a leading coefficient that is not to 1?
To factorise a trinomial (or Quadratic) expression, a different approach must be taken than the usual "what adds to the x-coefficient and multiplies to the constant".
As an example, let's use 4x2+ 10x + 4.
First, we check to see if the coefficients all have a common factor.
In this case, all of them can be divided by 2, so our expression becomes 2(2x2+ 5x + 2).
Next, we multiply the x2coefficient by the constant to get a "synthetic" constant. The constant is still 2, but we need to use this new, synthetic constant to continue.
2*2=4, so now we have 2(2x2+ 5x + 4).
Now, we find two numbers that add to 5 and multiply to this new synthetic constant 4: they are 4 and 1.
Now, we split the middle 5x term into two terms with coefficients 4 and 1, and we also bring back our old friend 2 as the original constant. This gives us:
2(2x2+ 4x + 1x + 2)
Next, we factorise the first and second terms together, and the third and fourth together. This gives us:
2( 2x(x+2) + 1(x+2) )
Then we tidy that up by putting the '2x' and the '1' into their own bracket, keeping it next to the (x+2) bracket to get:
2((2x+1)(x+2))
Remove the outermost set of brackets, as they serve no more purpose, to leave us with the final, fully factorised result:
4x2+ 10x + 4 = 2(2x+1)(x+2).
What else can I help you with?
In factoring a trinomial with a leading coefficient other than 1 what is the first step?
find a greatest common factor or GCFin factoring a trinomial with a leading coefficient other than 1 the first step is to look for a COMMON factor in each term
When factoring a trinomial is in the format Ax plus Bx plus C the number that is repersented by B should be viewed as the?
Ax + Bx + C is not a trinomial!
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
A trinomial with a leading coefficient of 3 and a constant term of -5?
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How do you factor a trinomial with a leading coefficient is equal to 1?
Suppose the trinomial is x2 + Bx + C You need to find a factor pair of C whose sum is B. If the factors are p and q (that is, pq = C and p+q = B), then the trinomial can be factorised as (x + p)*(x + q).
What is the leading coefficient of each fungtion?
what is the leading coefficient -3x+8
How do you find zeros when the leading coefficient is one?
The answer depends on the what the leading coefficient is of!
What is the difference in finding the m and n when factoring a polynomial with a leading coefficient that does not equal one?
The difference depends on what m and n equal. If they are both variable then it dpends on what the equations are for each variable.
What is leading coefficient?
It is the coefficient of the highest power of the variable in an expression.
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 quadratic equations already contains a perfect square trinomial?
A quadratic equation that contains a perfect square trinomial can be expressed in the form ( ax^2 + bx + c = 0 ), where the trinomial can be factored as ( (px + q)^2 ). This means that the equation can be written as ( a(px + q)^2 = 0 ), leading to solutions derived from ( px + q = 0 ). Examples include equations like ( x^2 + 6x + 9 = 0 ) or ( 4x^2 - 12x + 9 = 0 ). In these cases, the perfect square trinomial allows for straightforward factoring and finding of roots.
