Solve {l}{35a^3b}{-77a^2b}{+14ab}

$$\left. \begin{array} { l } { 35 a ^ { 3 } b } \\ { - 77 a ^ { 2 } b } \\ { + 14 a b } \end{array} \right.$$

Factor

$$7ab\left(a-2\right)\left(5a-1\right)$$

Solution Steps

Factor out 7.

$$7\left(5a^{3}b-11a^{2}b+2ab\right)$$

Consider 5a^{3}b-11a^{2}b+2ab. Factor out ab.

$$ab\left(5a^{2}-11a+2\right)$$

Consider 5a^{2}-11a+2. Factor the expression by grouping. First, the expression needs to be rewritten as 5a^{2}+pa+qa+2. To find p and q, set up a system to be solved.

$$p+q=-11 pq=5\times 2=10$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -11.

$$p=-10 q=-1$$

Rewrite 5a^{2}-11a+2 as \left(5a^{2}-10a\right)+\left(-a+2\right).

$$\left(5a^{2}-10a\right)+\left(-a+2\right)$$

Factor out 5a in the first and -1 in the second group.

$$5a\left(a-2\right)-\left(a-2\right)$$

Factor out common term a-2 by using distributive property.

$$\left(a-2\right)\left(5a-1\right)$$

Rewrite the complete factored expression.

$$7ab\left(a-2\right)\left(5a-1\right)$$

Evaluate

$$7ab\left(a-2\right)\left(5a-1\right)$$