Question

Pls help me with d others..pls need help asap...tia

Answer 1

d. d(2d-1)(3d+1)

e. (2e-3)(e-1)

f. 2(f-1)(2f-1)

Explanation:

d. First of all, factor out d. Then you have

d(6d² -d -1)

As always with these things, you're looking for factors of ac=(6)(-1) that sum to b=-1. These are -3 and 2, so you can write the quadratic factor as ...

6d² -3d +2d -1

and factor the pairs.

(6d² -3d) +(2d -1) = 3d(2d -1) +1(2d -1) = (3d+1)(2d-1)

So, the factorization of the given expression is ...

d(2d -1)(3d +1)

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e. In this one, you're looking for factors of 2·3 = 6 that sum to -5. Those are -2 and -3, so you have ...

2e² -5e +3 = 2e² -2e -3e +3 = 2e(e -1) -3(e -1) = (2e -3)(e -1)

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f. All the coefficients are even, so you can immediately factor out 2.

2(2f² -3f +1)

This requires you find factors of 2·1 = 2 that sum to -3. Those are -2 and -1, so you can write ...

2f² -f -2f +1 = f(2f -1) -1(2f -1) = (f -1)(2f -1)

so the factorization of the given expression is ...

2(f -1)(2f -1)

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Comment on your work

It looks like you have a process that works for you. I suggest keeping the signs of the variables positive where possible, so minus signs don't cause you confusion.

I also suggest you choose a method of displaying your answer that doesn't involve the use of what looks like a minus sign. I find it difficult to tell the difference between ...

• ans:-3 . . . . . an answer of +3 displayed in your style
• ans: -3 . . . . an answer of -3 seen by someone reading your style

Answer 2

see explanation

Explanation:

a

a² - 2a - 35

Consider the factors of the constant term (- 35) which sum to give the coefficient of the a- term (- 2)

The factors are - 7 and + 5

a² - 2a - 35 = (a - 7)(a + 5)

b

2b² + 8b - 42 ( take out a common factor of 2 from each term )

= 2(b² + 4b - 21)

Consider the factors of the constant term (- 21) which sum to give the coefficient of the b- term (+ 4)

The factors are + 7 and - 3

2b² + 8b - 42 = 2(b + 7)(b - 3)

c

2c² + c - 6

Consider the factors of the product of the c² term and the constant term which sum to give the coefficient of the c- term

product = 2 × - 6 = - 12, sum = + 1

The factors are + 4 and - 3

Split the middle term using these factors

2c² + 4c - 3c - 6 ( factor the first/second and third/fourth terms )

= 2c(c + 2) - 3(c + 2) ← factor out (c + 2)

= (c + 2)(2c - 3)

d

6d³ - d² - d ( factor out d from each term )

= d(6d² - d - 1)

Consider the factors of the product of the d² term and the constant term that sum to give the coefficient of the d- term

product = 6 × - 1 = 6, sum = - 1

The factors are - 3 and + 2

6d² - d - 1

= 6d² - 3d + 2d - 1

= 3d(2d - 1) + 1(2d - 1)

= (2d - 1)(3d + 1)

6d³ - d² - d = d(2d - 1)(3d + 1)

e

2e² - 5e + 3

product = 2 × 3 = 6, sum = - 5

Factors are - 2 and - 3

=2e² - 2e - 3e + 3

= 2e(e - 1) - 3(e - 1)

= (e - 1)(2e - 3)

f

4f² - 6f + 2 ( factor out 2 from each term )

= 2(2f² - 3f + 1)

product = 2 × 1 = 2, sum = - 3

Factors are -2 and - 1

2f² - 2f - f + 1

=2f(f - 1) - 1(f - 1)

= (f - 1)(2f - 1)

4f² - 6f + 2 = 2(f - 1)(2f - 1)