Solve 18x^4-42x^3+24x^2?

$$18 x ^ { 4 } - 42 x ^ { 3 } + 24 x ^ { 2 } ?$$

Factor

$$6\left(x-1\right)\left(3x-4\right)x^{2}$$

Solution Steps

Factor out 6.

$$6\left(3x^{4}-7x^{3}+4x^{2}\right)$$

Consider 3x^{4}-7x^{3}+4x^{2}. Factor out x^{2}.

$$x^{2}\left(3x^{2}-7x+4\right)$$

Consider 3x^{2}-7x+4. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-7 ab=3\times 4=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite 3x^{2}-7x+4 as \left(3x^{2}-4x\right)+\left(-3x+4\right).

$$\left(3x^{2}-4x\right)+\left(-3x+4\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(3x-4\right)-\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(x-1\right)$$

Rewrite the complete factored expression.

$$6x^{2}\left(3x-4\right)\left(x-1\right)$$

Evaluate

$$6\left(x-1\right)\left(3x-4\right)x^{2}$$