Question

Which of the binomials below is a factor of this trinomial? URGENT!!!

Answer 1

Answer:

C

Step-by-step explanation:

10×-28=-280

35-8=27

35×(-8)=-280

10x²+27x-28

=10x²+(35-8) x-28

=10x²+35x-8x-28

=5x(2x+7)-4(2x+7)

=(2x+7)(5x-4)

Answer 2

Answer: C) 2x+7

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Explanation:

One way we can factor is through use the of the quadratic formula.

Let $10x^2+27x-28 = 0$

For now, the goal is to find the two roots of that equation.

Plug a = 10, b = 27, c = -28 into the quadratic formula

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(27)\pm\sqrt{(27)^2-4(10)(-28)}}{2(10)}\\\\x = \frac{-27\pm\sqrt{1849}}{20}\\\\x = \frac{-27\pm43}{20}\\\\x = \frac{-27+43}{20} \ \text{ or } \ x = \frac{-27-43}{20}\\\\x = \frac{16}{20} \ \text{ or } \ x = \frac{-70}{20}\\\\x = \frac{4}{5} \ \text{ or } \ x = -\frac{7}{2}$

The two roots are x = 4/5 and x = -7/2

For each root, rearrange the equation so we have 0 on the right hand side, and it's ideal to get rid of the fractions

x = 4/5

5x = 4

5x-4 = 0 gives us one factor

and

x = -7/2

2x = -7

2x+7 = 0 gives the other factor

The two factors are 5x-4 and 2x+7

Note how (5x-4)(2x+7) = 0 leads to the two separate equations of 5x-4 = 0 and 2x+7 = 0 due to the zero product property. Solving each individual equation leads to the two roots we found earlier.

Alternative methods to solve this problem are the AC factoring method (which leads to factor by grouping), using the box method, or you could use guess and check.