PLease help me with factorising...?

Question

(Question)1. The function is: g(x)=2x^2-7x+3
Factorised form:

Note: I can't get it right! I was so close I thought but it was +2 instead of +3. The rest of it was correct. Please help me put the function into factorised form.
How do you work it out other than the guess and check method which is time consuming and at times frustrating?

(Question)2. Function: 3x^2-7x-6
Factorised form:

Note: I have tried this one as well. I can't work it out!

I also have a question which asks me to factorise a function completely. Is this different to putting the function in factorised form? Is it just different wording of the same question really?

3. Is there any relationship between the zeros and the coefficient of a function? How are they related?

Please help me out Math whizzes! Thankyou so much.

Answer 1

2x^2 - 7x + 3
since the middle term is negative and the constant positive both factors will have the form (ax - b). 2 is prime, so you have (2x - a)(x - b). 3 is also prime, so you are down to 2 choices for factors:
(2x - 3)(x - 1) or (2x - 1)(x - 3). The first doesn't work because -2x - 3x = -5x, but - 6x - 1x = - 7x, which is the term you want, so the second is the solution:
2x^2 - 7x + 3 = (2x - 1)(x - 3)

3x^2 - 7x - 6 works out in about the same way, except now 6 can be formed by 1*6 or 2*3 and the 6 is negative, so you need a form of (ax - b)(cx + d), so you need products whose difference is 7
3*6 - 1*1 = 17, nope
3*3 - 1*2 = 7, so now all we need to do is get the sign right. the 3*3 must end up negative, and the 1*2 must end up positive.
(3x + 2)(x - 3) works nicely.

Each zeros of a quadratic function is going to be some factor of the constant divided by some factor of the squared term. From the above, setting each term = 0,
x = -2/3 or x = 3/1

Answer 2

The answers are:
2x^2 - 7x + 3 = (2x-1)(x-3)
3x^2 - 7x - 6 = (3x+2)(x-3)

Now I'll walk you through it...

What you need to do is first look at the number by itself and the number in front of the x^2 and determine the possible factors of them.

(For #1)
2 = 2*1 (could be -2*-1, but we don't usually use negatives for the first term.)
3 = 3*1 or -1*-3

Note that the 3 is positive, which means that those factors are either both positive or both negative. Since the middle term is -7 (negative), the factors of 3 both need to be negative.

So, your choices are:
(2x-1)(x-3) and (2x-3)(x-1)
Now, it's kind of guess-and-check... which of those two combinations gives you -7x in the middle?

(for #2)
The factors are:
3 = 3*1
-6 = -1*6, 1*-6, -2*3, 2*-3

This one is a bit harder because you have more factors of 6 to work with. Again, the middle term is -7, so we need some combination to get that. First, put in what you do know:

(3x ________)(x _________)

If we use a combination of the 6*1 (with one of them negative), we end up with one term of 3*6 and 1*1, or 18 and 1, which you can't use to get to -7 with at all.
The other option with 6*1 is 3*1 and 6*1. Now you're working with the numbers 3 and 6, which you can't add or subtract to get to -7.

So we know, we need some combination of 2*3.
So we either have 2*3 and 3*1 (6 and 3), or we have 3*3 and 2*1 (9 and 2). Again 6 & 3 won't get us to 7, but 9 & 2 will!

Now we know:
(3x ___ 2)(x ___3)

Now we just have to figure out which is negative... we want -7, so we need -9 + 2 to get there. So it becomes:
(3x+2)(x-3)
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Factorize complete = put in factorized form. It's just telling you to make sure that you factor everything possible. ex: 2x+4 is not factored completely... 2(x+2) is.
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3) I don't believe there's a relationship between the coefficent and the zeros. The number of zeros is somewhat determined by the degree of the equation (the highest exponent). The TYPE of zeros depends on a combination of things - it does rely on the ALL the coefficients listed in the function, not just the first one.

Answer 3

Factorising completely= puttng in factorised form. both mean the same

Rule for this type of factors (ax^2 +bx + c):
split b into p + q so that pq = ac; p+q=b

eg 1:
2x^2-7x+3
ac = 2(3)= 6
b = 7 = 6 + 1
pq = 6(1)= 6
rewriting,
2x^2 - 6x - x + 3
2x(x-3) -1(x-3)
(x-3)(2x - 1) Ans


eg 2:
3x^2-7x-6
ac = 3(-6)= 18
b = -7 = 2 - 9
pq = 2(9) = 18
rewriting,
3x^2 -9x +2x -6
3x(x-3) +2 (x-3)
(x-3)(3x+2) Ans

Answer 4

g(x)=2 x^2 - 7 x + 3
If you solve the cuadratic equation g(x)=0, you obtain as roots

x = [7 +- SquareRoot(7^2-4*2*3)] / 2*2 = (7+- 5)/4, so x1= 3, x2=1/2

Then g(x)=0 factorizes as (x-3)*(x-1/2)*2 (we multiply by 2 to avoid fractions). We get
g(x)=(x -3)(2x-1)


Same procedure for the second function.
Roots of the equation are (7+-11)/6, x1= 3, x2= -2/3, so we have
g(x) = (x-3)(x+2/3)*3 = (x-3)(3x+2)


3) Of course the zeroes of a function depend on their coefficients. The number of zeroes does not depend on its coefficiente, for it is equal to the degree of the equation (the highest exponent).

Answer 5

Question 1
2x² - 7x + 3 = (2x - 1)(x - 3)

Method
The factors of 2x² are 2x and x so 2x and x are placed in brackets as shown.

The factors of 3 are 3 and 1 or 1 and 3.
Choose 1 and 3 as this gives inner product of 1x and outer product of 6x
Then have to decide on signs and in order to obtain -7x require -1 and -3.

Question 2
3x² - 7x - 6 = (3x + 2)(x - 3)

As a check (3x) x (x) = 3x²
-9x + 2x = -7x
2 x - 3 = - 6

I would not become bogged down with wording of next question----just factorise it or post question on this site.

Question 3
Perhaps you could post a sample question that I could tackle?

Answer 6

Keisha, as the first poster says, you can use the quadratic formula for this. You will probably learn about this shortly in school. Formula is shown here

http://www.purplemath.com/modules/quadform.htm

For the first one, if you use the formula you will get
x = 3 and x = 0.5

Therefore 0 = x - 3

and

0 = x - 0.5
multiply both sides by 2, and you get
0 = 2x - 1

This translates to a factorized form of (x-3)(2x-1)

try this method for the second

Answer 7

to ascertain if an equation may nicely be factorised, calculate b^2-4ac the position b is the co-powerful of the potential-a million time period(4 hence), a is the co-powerful of potential-2 time period(6 hence), and c(10 hence) is the lone volume. 16-240=-224 If the reply comes out to be adverse, then the equation won't be able to be factorised.

Answer 8

Sure. For number one. Use the following format:

(ax+b)(cx+d)...That is the template for factoring when it comes to polymonials.

So
a*c=2 <--- ax+cx =2x^2, first part of your equation
ad+bc=-7 <---- ax*d + bx*c= -7, 2nd part of your equation
b*d=3 <---- +b*+d = +3, last part of your equation

Any number'll work for the a*c. Well, you know that a will be used to make 7, which is a negative number.

a=-1
c=-2

ac=2

The first part of your equation fits so far.

2nd part: ad+bc=-7....

Enter the solved= -1d+b(-2)=-7.... now, take any simple factor of addition [since it requires addition for this equation, as you see above after 'enter the solved']...Let's take -1-6 [stick to 1s]

d=1
b=3

There, now the whole equation is finished. =)

Plug in according to (ax+b)(cx+d)

(-x+3)(-2x+1) <--- D= CONGRATS!

As for the next one...

3x^2-7x-6

(ax+b)(cx+d)...That is the template for factoring when it comes to polymonials.

So
a*c=3 <--- ax+cx =3x^2, first part of your equation
ad+bc=-7 <---- ax*d + bx*c= -7, 2nd part of your equation
b*d=6 <---- +b*+d = +6, last part of your equation

Any number'll work for the a*c. Well, you know that a will be used to make 7, which is a negative number, once again. [Some guessing/logic application will be involved]

a=-1
c=-3

ac=3

The first part of your equation fits so far.

2nd part: ad+bc=-7....

Enter the solved= -1d+b(-3)=-7.... now, take any simple factor of addition [since it requires addition for this equation, as you see above after 'enter the solved']...Let's take - 1 - 6 = -7 [stick to 1s]

d=1
b=2

There, now the whole equation is finished. =)

Plug in according to (ax+b)(cx+d)

(-x+3)(-2x+1) <--- D= CONGRATS!

>>> The question that asks you to factorize a function completely. The problem requires different steps to factorize.

Usually 3-term polynomials, like the one above, use the (ax+b)(cx+d) template. 4-term polynomials are usually asked to "factor completely" No it is not different from putting it into factorized form, it is in fact, putting the function into factorized form.

However, insead of (ax+b)(cx+d), factoring completely would be ax(x^2-b)(cx+d), a different form.

3. Yes there is relation between the zeros and coefficients of a function. The zeros apply to, and will be added on/multiplied with the coeffecients and, indirectly, by changing the coefficient, will change the multiplified amount of the variable.

Answer 9

Heres help for the 1st question. you see that you need 2 numbers that will multiply to give +3, clearly only 3 and 1 will do that. secondly you look at the x term. you see that you need -7 of them. obviously there are (-3 x 2 )+ -1 x's. in facotrized form, therefore it is (2x-1)(x-3)

Answer 10

x = [-b (+/-) (b²-4ac)^0.5 ]/2
where the function is a²x+bx+c=0

u'll get 2 answers for x
x1, and x2

then.. in the end.. u can factorize with
g(x) = (x-x1)(x-x2)

am i making any sense? i guess it is hard to type the formula here. if no one else gives u a better answer, or you want to know what i mean, send me an email, and i'll type it out for you on a word document.