How do you factor the trinomial $4 y^{4} x + 14 y x^{2} + 10 y^{2} x^{2} - 8 x y$ $4y^4x + 14yx^2 + 10y^2x^2 - 8xy$ ?

Question

How do you factor the trinomial $4 y^{4} x + 14 y x^{2} + 10 y^{2} x^{2} - 8 x y$ $4y^4x + 14yx^2 + 10y^2x^2 - 8xy$ ?

1 Answer

Impact of this question

1532 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-12T16:59:10

$4 y^{4} x + 14 y x^{2} + 10 y^{2} x^{2} - 8 x y = 2 y x (2 y^{3} + 7 x + 5 y x - 4)$ $4y^4x+14yx^2+10y^2x^2-8xy=2yx(2y^3+7x+5yx-4)$

with no simpler factorisation.

Explanation:

This is a quadrinomial not a trinomial.

First separate out the common factor $2 y x$ $2yx$ :

$4 y^{4} x + 14 y x^{2} + 10 y^{2} x^{2} - 8 x y = 2 y x (2 y^{3} + 7 x + 5 y x - 4)$ $4y^4x+14yx^2+10y^2x^2-8xy=2yx(2y^3+7x+5yx-4)$

I suspect this is as far as we can factor it, but let's try.

Since the leading term is $2 y^{3}$ $2y^3$ and all other terms are of lower degree, then if $(2 y^{3} + 7 x + 5 y x - 4)$ $(2y^3+7x+5yx-4)$ splits into two factors then one is linear and the other is quadratic:

$2 y^{3} + 7 x + 5 y x - 4$ $2y^3+7x+5yx-4$

$= (2 y + a x + b) (y^{2} + c y x + {d x}^{2} + e y + f x + g)$ $=(2y+ax+b)(y^2+cyx+dx^2+ey+fx+g)$

$= 2 y^{3} + (a + 2 c) y^{2} x + a {d x}^{3} + (b + 2 e) y^{2} + (a c + 2 d) y x^{2} + (b c + a e + 2 f) y x + (b e + 2 g) y + (b d + a f) x^{2} + (b f + a g) x + b g$ $=2y^3+(a+2c)y^2x+adx^3+(b+2e)y^2+(ac+2d)yx^2+(bc+ae+2f)yx+(be+2g)y+(bd+af)x^2+(bf+ag)x+bg$

Hence:

(i) $a + 2 c = 0$ $a+2c=0$

(ii) $a d = 0$ $ad=0$

(iii) $b + 2 e = 0$ $b+2e=0$

(iv) $a c + 2 d = 0$ $ac+2d=0$

(v) $b c + a e + 2 f = 5$ $bc+ae+2f=5$

(vi) $b e + 2 g = 0$ $be+2g=0$

(vii) $b d + a f = 0$ $bd+af=0$

(viii) $b f + a g = 7$ $bf+ag=7$

(ix) $b g = - 4$ $bg=-4$

From (ii) at least one of $a = 0$ $a=0$ or $d = 0$ $d=0$ .

If $a = 0$ $a=0$ then from (iv) we find $d = 0$ $d=0$ .

So we at least know $d = 0$ $d=0$

Since $d = 0$ $d=0$ , then from (iv) we find $a c = 0$ $ac=0$ , so at least one of $a = 0$ $a=0$ or $c = 0$ $c=0$ . Then from (i) we find that the other is zero too.

So given that $a = c = d = 0$ $a = c = d = 0$ , the remaining equations become:

(iii) $b + 2 e = 0$ $b+2e=0$

(v) $2 f = 5$ $2f=5$

(vi) $b e + 2 g = 0$ $be+2g=0$

(viii) $b f = 7$ $bf=7$

(ix) $b g = - 4$ $bg=-4$

From (v) $f = \frac{5}{2}$ $f = 5/2$

From (viii) $b = \frac{7}{f} = \frac{14}{5}$ $b = 7/f = 14/5$

From (ix) $g = - \frac{4}{b} = - \frac{20}{14} = - \frac{10}{7}$ $g = -4/b = -20/14 = -10/7$

From (iii) $e = - \frac{b}{2} = - \frac{7}{5}$ $e = -b/2 = -7/5$

So:

$b e + 2 g = \frac{14}{5} \cdot (- \frac{7}{5}) + 2 (- \frac{10}{7})$ $be+2g = 14/5*(-7/5)+2(-10/7)$

$= - \frac{98}{25} - \frac{20}{7} \neq 0$ $= -98/25-20/7 != 0$ violating (vi)

So there is no simpler factorisation.

Science

Math

Humanities

... and beyond

How do you factor the trinomial $4 y^{4} x + 14 y x^{2} + 10 y^{2} x^{2} - 8 x y$ $4y^4x + 14yx^2 + 10y^2x^2 - 8xy$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor the trinomial 4y4x+14yx2+10y2x2−8xy4y^4x + 14yx^2 + 10y^2x^2 - 8xy?

1 Answer

Explanation:

Related questions

Impact of this question

How do you factor the trinomial $4 y^{4} x + 14 y x^{2} + 10 y^{2} x^{2} - 8 x y$ $4y^4x + 14yx^2 + 10y^2x^2 - 8xy$ ?