Can You Factorize These Expressions?

Question

1. x squared + 3x - 70

2. x squared + 4x + 3

3. x squared + 9x + 8

4. x squared - 6x + 9

5. x squared - 2x - 24

6. x squared + 7x + 10

Answer 1

1) (x+10)(x-7)
2) (x+1)(x+3)
3) (x+1)(x+8)
4) (x-3)(x-3) = (x-3)²
5) (x-6)(x+4)
6) (x+5)(x+2)

Answer 2

I will show you one of them using the method of completing the square.

x^2+ 4x + 3
First we need the roots, that is when is the equation = 0:

x^2 + 4x + 3 = 0

x^2 + 4x = -3 //add 3 each side to balanced equation

x^2 + 4x +4 -4 = -3 // add and subtract half the coefficient of x to left hand side. Coeffiecient of x is 4, half 4 is 2, 2 squared is 4. This does not chang the equation as 4-4 = 0 and you can add 0 without changing.

Look carefully at x^2 + 4x +4, it is made up of (x + 2)^2. if you take X^2 + ax you can rewrite this as x^2 + ax + (a/2)^2 - (a/2)^2, where a is the coefficient of x. Then x^2 + ax + (a/2)^2 - (a/2)^2 can be written as (x + a/2)^2 - (a/2)^2

Returning to your example a = 4 so we have

(x + 2)^2 -4 = -3

(x + 2)^2 = 1 //add 4 each side

x + 2 = +1, -1 //take square root of each side

Case +1:
x + 2 = +1
x = -1 //subtract 2 each side, this is a root.
x + 1 = 0 //add 1 each side to get 0 on right hand side, this is a factor

Case -1:
x + 2 = -1
x = -3 //subtract 2 each side, this is a root.
x + 3 = 0 // add three each side to get 0 on right hand side, this is a factor

so the factors are (x + 1) and (x + 3)

This always works and in my opinion if you prqactice them you will get very fast and it beats trying to guess and try.

Once you get the hang of it you can shortcut the two cases:

Case +1:
x + 2 = +1
x + 1 = 0 //subtract 1 each side to get 0 on right hand side, this is a factor

Case -1:
x + 2 = -1
x + 3 = 0 // add one each side to get 0 on right hand side, this is a factor

Peter

Answer 3

1.
(x - 7)(x + 10)

2.
(x + 3)(x + 1)

3.
(x + 8)(x + 1)

4.
(x - 3)(x - 3)
= (x - 3)^2

5.
(x - 6)(x + 4)

6.
(x + 5)(x + 2)

Answer 4

Use the substitution rule of s = x^2 then you definately've the quadratic : 2s^2 + 14s + 24 = 0 it truly is decreased to : s^2 + 7s + 12 = 0 you may now element this into : (s + 3)(s + 4) So the aspects are truly the below, once you positioned the x^2 back in the place the s is : (x^2 + 3)(x^2 + 4) or in case you have no longer set it to 0 : (2)(x^2 + 3)(x^2 + 4) till you desire the certainly roots of this (which would be imaginary numbers related to the iota), it truly is your answer. the certainly roots of this are : x = i?3, -i?3, 2i, - 2i

Answer 5

1.-----(x + 10).(x - 7)
2.-----(x + 3).(x + 1)
3.-----(x + 8).(x + 1)
4.-----(x - 3).(x - 3)
5.-----(x - 6).(x + 4)
6.-----(x + 5).(x + 2)

Answer 6

Having the answers given to you is about as much use as a glass hammer if you don't understand how to factorize.

Answer 7

Yes indeed I can factorize them, thanks! Factorizing is very easy.

Just find things which, when multiplied together, equal what you have to start with. For example:

Question: factorize 6.
Answer: 3x2.

Learn how to do it yourself instead of cheating on your homework. There's no shame in asking your teacher for extra help.