Solve 2a^2-4a-5a+10

$$2 a ^ { 2 } - 4 a - 5 a + 10$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$2a^{2}-9a+10$$

Factor the expression by grouping. First, the expression needs to be rewritten as 2a^{2}+pa+qa+10. To find p and q, set up a system to be solved.

$$p+q=-9 pq=2\times 10=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$p=-5 q=-4$$

Rewrite 2a^{2}-9a+10 as \left(2a^{2}-5a\right)+\left(-4a+10\right).

$$\left(2a^{2}-5a\right)+\left(-4a+10\right)$$

Factor out a in the first and -2 in the second group.

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

Factor out common term 2a-5 by using distributive property.

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

Evaluate

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

Solution Steps

Combine -4a and -5a to get -9a.

$$2a^{2}-9a+10$$