Question

How do you factor these questions?

Question 1: 20(exponent 2) - 13 (exponent 3)
Question 2: (x+3)(exponent 2) + 5(x+3) +6
Question 3: 2(x-d)(exponent 2) + 9(x-d) +10

Answer 1

factoring an expression is nothing but expressing the thing in a multiplication form

question 1-factor \(20^2-13^2\)

apply this identity-> \(\large \color{orangered}{a^2-b^2=(a+b)(a-b)}\)

Answer 2

question 2- factor \((x+3)^2+5(x+3)+6\)

use the method of splitting the middle term

hint:- you can write the given expression like this-
\((x+3)^2+3(x+3)+2(x+3)+6\)

now try to make factors by taking common things out

\(\color{red}{take~2~ common\\
~~~~~~~~~~from~ this}\)
\(\underbrace{(x+3)^2+3(x+3)}+\overbrace{2(x+3)+6}\)
\(\color{blue}{take~ (x+3)~ common\\
~~~~~~from~ this}\)

-> \((x+3)(x+3+3)+(2)(x+3+3)\)
-> \(\underbrace{(x+3)(x+6)+(2)(x+6)}\)
\(\color{blueviolet}{take~(x+6)~common ~from~here}\)

-> \((x+6)(x+3+2)\)
->\((x+6)(x+5)\)

Answer 3

question 3-> factor \(2(x-d)^2+9(x-d)+10\)

again use splitting the middle term method

-> \(2(x-d)^2+5(x-d) +4(x-d)+10\)

->\((x-d)(2(x-d)+5)+2(2(x-d)+5)\)

-> \((2(x-d)+5)(x-d+2)\)

simplify it from here :)

Answer 4

Thank you very much but I'm just a little but confused about simplifying (2(x-d) + 5)(x-d+2) do I use distributive property? @imqwerty

Answer 5

here we are not distributing anything

what we do is->

\(\underbrace{(x-d)\color{orangered}{(2(x-d)+5)}+2\color{orangered}{(2(x-d)+5)}}\)
\(\color{hotpink}{take~(2(x-d)+5) ~common~from~these~two ~terms}\)

you get this->

\(\color{orangered}{(2(x-d)+5)}((x-d)+2)\)

Answer 6

well yes
in a way we can say that we used distributive property
because distributive property says that-

\(a(b+c)=ab+ac\)

Answer 7

Ohh I see thank you very much

Answer 8

np :)