Question

Write 12h^2-31h+20 as the product of two factors. What are the zeros of the related function?

Answer 1

Split -31 as -16 - 15

Answer 2

12h^2(-16-15)h+20? How do you rewrite that

Answer 3

Factoring a trinomial of the form \(ax^2 + bx + c\), where \(a\ne 1\)

1. First try to factor a common term out of all factors.
2. Multiply ac together.
3. Find two numbers that multiply to ac and add to b. Call these numbers p and q.
4. Split up the middle term, bx, into px + qx.
5. Factor by grouping.

Answer 4

Step 1.
12, -31, and 20 have no common factors, so we cannot factor out a common factor.

Step 2.
ac = 12 * 20 = 240

Step 3.
We need 2 numbers that multiply to 240 and add to -31.
@ganeshie8 alreay supplied the numbers. They are -15 and -16.

Step 4.
Split up the bh term term into two terms using the numbers -15 and -16.
\(12h^2 - 15 h - 16h + 20\)

Step 5.
Factor by grouping.
That means, take out a common factor out of the first two terms.
Take out a common factor out of the last two terms.
Then take out a common factor (a binomial).

\(12h^2 - 15h - 16h + 20\)

\(= 3h(4h - 5) - 4(4h - 5) \)

\(= (4h - 5)(3h - 4) \)

Answer 5

Thank yyouu :)

Answer 6

For the second part of the questions, "what are the zeros of the related function?" What exactly is that asking for?

Answer 7

@mathstudent55

Answer 8

They want you to set that polynomial equal to zero and solve it as an equation.

Answer 9

Since we already factored it, set each factor equal to zero, and solve it for h.
4h - 5 = 0 or 3h - 4 = 0
Solve both equations above and give the answers separated by the word "or".