MHB How can I factorize this polynomial?

AI Thread Summary
The discussion focuses on factorizing the polynomial expression 6a² - 3ab - 11ac + 12ad - 18b² + 36bc - 45bd - 10c² + 27cd - 18d². A proposed factorized form is (Aa + Bb + Cc + Dd)(Wa + Xb + Yc + Zd), and various approaches to decomposition are explored. The user successfully identifies factors for specific parts of the polynomial, such as (2a - 5c)(3a + 2c) and (3a - 6b)(2a + 3b). There is also discussion about potential sign errors and strategies for selecting which trinomial to factor first. The conversation emphasizes the complexity of polynomial factorization and the need for careful comparison of terms.
bergausstein
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Decompose

$$6a^2-3ab-11ac+12ad-18b^2+36bc-45bd-10c^2+27cd-18d^2$$

I noticed that the factorized form would be $$(Aa+Bb+Cc+Dd)(Wa + Xb + Yc + Zd)$$

Which is similar to the factorized form $$(Aa+Bb+Cc)(Wa+Xb+Yc)$$

$$Yc(Aa+Bb)+Cc(Wa+Xb) = c(CX+BY)$$

Is there a way that I can somehow use this to decompose the original polynomial expression? I'm stuck. I need help.
 
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bergausstein said:
Decompose

$$6a^2-3ab-11ac+12ad-18b^2+36bc-45bd-10c^2+27cd-18d^2$$

I noticed that the factorized form would be $$(Aa+Bb+Cc+Dd)(Wa + Xb + Yc + Zd)$$

Which is similar to the factorized form $$(Aa+Bb+Cc)(Wa+Xb+Yc)$$

$$Yc(Aa+Bb)+Cc(Wa+Xb) = c(CX+BY)$$

Is there a way that I can somehow use this to decompose the original polynomial expression? I'm stuck. I need help.

If from the above we take (Aa+Bb)(Wa+Xb) we get quadratic polynomial in ab so one

let us see for a c because coefficient of $^2$, $c^2$ and $ac$ do not have any constant as a common factor

we get $6a^2-11ac - 10c^2 = 6a^2-15ac + 4ac = 10c^2 = 3a(2a-5s) + 2c(2a-5s) = (2a-5c)(3a + 2c)$

so we get factor of the form(2a+Ab-5c +Bd)(3a+Cb+2c + Dd)

taking $6a^2-3ab -18b^2 = 3(2a^2 - ab - 6b^2) = 3(2a+3b)(a-2b) = (3a - 6b)(2a+3b)$

comparing above we get A = 3 and C = 6

so we get factor (2a+3b-5c +Bd)(3a+6b+2c + Dd)

now take trinoomal of a d

$6a^2 +12ad-18d^2 = 6(a^2+2ad-3d^2)= 6(a+3d)(a-d) = (2a+6d)(3a-3d)$

comparing with above we have B= 6 and D = -3 and hence facfored as = (2a+3b-5c +6d)(3a+6b+2c -3d)
you can multiply and check it out.
 
kaliprasad said:
If from the above we take (Aa+Bb)(Wa+Xb) we get quadratic polynomial in ab so one

let us see for a c because coefficient of $^2$, $c^2$ and $ac$ do not have any constant as a common factor

we get $6a^2-11ac - 10c^2 = 6a^2-15ac + 4ac = 10c^2 = 3a(2a-5s) + 2c(2a-5s) = (2a-5c)(3a + 2c)$

so we get factor of the form(2a+Ab-5c +Bd)(3a+Cb+2c + Dd)

taking $6a^2-3ab -18b^2 = 3(2a^2 - ab - 6b^2) = 3(2a+3b)(a-2b) = (3a - 6b)(2a+3b)$

comparing above we get A = 3 and C = 6

so we get factor (2a+3b-5c +Bd)(3a+6b+2c + Dd)

now take trinoomal of a d

$6a^2 +12ad-18d^2 = 6(a^2+2ad-3d^2)= 6(a+3d)(a-d) = (2a+6d)(3a-3d)$

comparing with above we have B= 6 and D = -3 and hence facfored as = (2a+3b-5c +6d)(3a+6b+2c -3d)
you can multiply and check it out.

Hi kaliprasad! It did check out. But I think you made a sign error $(3a-6b+2c-3d)$ instead of $(3a+6b+2c-3d)$. How do you decide which trinomial to take first? And what if I choose $-10c^2+27cd-18d^2$? And where do you compare it to after factoring?
 
Last edited:
bergausstein said:
Hi kaliprasad! It did check out. But I think you made a sign error $(3a-6b+2c-3d)$ instead of $(3a+6b+2c-3d)$. How do you decide which trinomial to take first? And what if I choose $-10c^2+27cd-18d^2$? And where do you compare it to after factoring?

you are right

taking 6a2−3ab−18b2=3(2a2−ab−6b2)=3(2a+3b)(a−2b)=(3a−6b)(2a+3b)6a2−3ab−18b2=3(2a2−ab−6b2)=3(2a+3b)(a−2b)=(3a−6b)(2a+3b)

comparing above we get A = 3 and C = 6

so we get factor (2a+3b-5c +Bd)(3a+6b+2c + Dd)

it should be

A = 3 and B= -6 hence your result
 
bergausstein said:
Hi kaliprasad! It did check out. But I think you made a sign error $(3a-6b+2c-3d)$ instead of $(3a+6b+2c-3d)$. How do you decide which trinomial to take first? And what if I choose $-10c^2+27cd-18d^2$? And where do you compare it to after factoring?

if you choose $-10c^2+27cd-18d^2$ you get
$-10c^2+15cd + 12cd - 18d^2= -5c(2c-3d) + 6d(2c-3d) = (2c-3d)(6d-5c) $

then taking combination of polynomial of b c or bd and so on
 
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