I need the process explained to me for three factoring problems.

a4b+a2b3, 18z+45+z2, 8x2-2xy-y2

I need the process explained to me for three factoring problems.

a4b+a2b3, 18z+45+z2, 8x2-2xy-y2

Tutors, please sign in to answer this question.

I think your first question is (((a^4)b)+((a^2)(b^3))).

According to the distributive property, (a^2)b can be taken out of the bracket from both the terms. (the two terms are ((a^4)b) and ((a^2)(b^3)).)

This will give you (a^2)b((a^2)+(b^2)) for your first question.

For your second question, I think you mean to ask 18z+45+(z^2). (this problem has three terms: 18z, 45, (z^2).) What I would do first thing would be to rearrange this problem like this (z^2)+18z+45.

To solve this problem we must find all the factor pairs of 45. (this is easy if you know your multiplication tables. If you don't know your multiplication tables then stop what you are doing and learn those first thing!) So we know the factor pairs of 45 are 1x45, 3x15, and 5x9. We need two factors of one pair that when added they will equal 18. 3 and 15 when added do equal 18, and when multiplied to equal 45.

Rearrange your problem so that it will look like this. (z^2)+15z+3z+45. (your 18z here is split between 15z and 3z). Now add two more brackets in your problem like this: ((z^2)+15z)+(3z+45)

Now using the distributive property take out the common factors in each bracket and keep doing that to get the answer. For the first bracket z is common and for the second bracket 3 is common. So you should get this: (z((z+15))+3((z+15))). Use the distributive property once more: z+15 is common for both the terms, so you should take it out of the bracket: (z+15)((z+3)).

For your third problem I think you mean it to be written like this (8x^2)-(2xy)-(y^2).

Note down the multiplication of 8 and -1, or "a" and "c" in this equation. You get -8.

Now the value "b" is -2. You need to find two numbers that multiply to -8 and add to -2. For doing this you need to find a factor pair with factors that add to -2. You get -4 and 2.

Rearrange your equation so that -4xy+2xy will fit in place of -2xy. So your equation will be (8x^2)-(4xy)+(2xy)-(y^2). Put two extra brackets in the problem so that it will look like this. ((8x^2)-(4xy))+((2xy)-(y^2)). Use the distributive property to distribute common factors in both the brackets. Keep doing this to get the answer. for the first bracket the common factor is 4x and for the second bracket the common factor is y. So your next step will look like this: 4x(2x-y)+y(2x-y) Now use the distributive property again and your common term is 2x-y so take that out of the brackets. (2x-y)((4x+y)). That's your answer.

I hope this helps. I tried my best explaining it online.

1. a4b+a2b3 = a2b(a2+b2) a2b is taken common from each term

2. 18z+45+z2 = z2+18z+45 = z2+15z+3z+45 = z(z+15)+3(z+15) = (z+15)(z+3) ... 15z+3z = 18z and 15*3 = 45

3. 8x2-2xy-y2 = 8x2-4xy+2xy-y2 = 4x(2x-y)+y(2x-y) = (2x-y)(4x+y) ...-4xy+2xy = -2xy and 4*2 = 8

1. we can factorize the first question a^{4 }b +a^{2} b^{2
} as

a^{4}b + a^{2}b^{2}= a^{2} b( a^{2}+b^{2})--------- because a^{2
}b is common for both terms

2. The second equation 18z +45+z^{2 } can be factorized as follows:

to factorize z^{2 }+ 18z +45 we need to find two numbers with their sum 18 and their product 45;

these two numbers are 15 and 3

Hence we can write z^{2 }+18z +45 as z^{2 }+ 18z +45 = z^{2
}+15z +3z+45

the we use factorization by grouping z^{2}+15z +3z +45= z(z +15) +3(z +15)

= (z+15)(z +3) is the answer

3. To factorize 8x^{2} -2xy-y^{2 }we need to find two numbers with sum -2 and product -8

these two numbers are -4 and 2

so we can write 8x^{2} -2xy -y^{2} = 8x^{2 }-4xy +2xy -y^{2}

= 4x(2x-y)+ y(2x-y)

= (4x+y)(2x-y) is the factorization

Balcha Borgi

Jonathan S.

Certified, Experienced Tutor and Teacher – Physics, Math, SATs

Brooklyn, NY

4.8
(92 ratings)

Amaan M.

Math/Economics Teacher

New York, NY

5.0
(68 ratings)

Dana R.

English/Math SAT PrepTutor with PhD

Lynbrook, NY

4.8
(85 ratings)