MHB How can I factorize this polynomial?

[bergausstein]

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.

 

[kaliprasad]

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.

 

[bergausstein]

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?

 

[kaliprasad]

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

 

[kaliprasad]

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