How do you factor the quadratic expression completely? #2x^2 - 13x + 20#

Question

Eva Abramson · Accepted Answer

#(x-4)(2x-5)#

#2x^{2}-13x+20#

Factors of #20 * 2# which add up to -13 are -5 and -8

so you can replace -13 with -5 and -8 such that:

#2x^{2}-5x-8x+20# which goes to: #x(2x-5)-4(2x-5)#

Take the expressions not in the brackets together to get: #(x-4)(2x-5)#

Nathan Axel · Answer

#(x-4)(x-5/2)#

#2x^2-13x+20#

The simplest way to do this is using the quadratic equation:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Where, #a=2# #b=-13# #c=20#

#x=(-(-13)+-sqrt((-13)^2-4*2*20))/(2*2)#

#x=((13)+-sqrt(169-160))/(4)# #x=((13)+-sqrt(9))/(4)# #x=4# and #x=5/2#

#x-4=0# #x-5/2=0#

Therefore,

#2x^2-13x+20= (x-4)(x-5/2)#

Hazel Anglin · Answer

undefined To factor the quadratic expression $2x^2 - 13x + 20$ completely, you can use the factoring method. First, find two numbers that multiply to the product of the leading coefficient (2) and the constant term (20), which is $2 \times 20 = 40$, and add up to the coefficient of the linear term (-13). The numbers are -8 and -5. Rewrite the middle term as the sum of these two numbers:

$$2x^2 - 8x - 5x + 20$$

Next, factor by grouping:

$$2x(x - 4) - 5(x - 4)$$

Now, notice that both terms have a common factor of $x - 4$. Factor this out:

$$(2x - 5)(x - 4)$$

So, the quadratic expression $2x^2 - 13x + 20$ factors completely as $(2x - 5)(x - 4)$.