Math  /  Algebra

QuestionFill in the gaps to factorise this expression. x2+6x27=(x(x+)x^{2}+6 x-27=(x-\not)(x+\ldots)

Studdy Solution

STEP 1

1. We are given a quadratic expression x2+6x27x^2 + 6x - 27 to factor.
2. The expression can be factored into the form (xa)(x+b)(x - a)(x + b).
3. We need to find the values of aa and bb such that the expression is equivalent to the original quadratic.

STEP 2

1. Identify the values of aa and bb that satisfy the conditions for factoring a quadratic expression.
2. Verify the factorization by expanding the factored form to ensure it matches the original expression.

STEP 3

Identify the values of aa and bb such that:
ab=27 a \cdot b = -27 a+b=6 -a + b = 6
We need two numbers whose product is 27-27 and whose sum is 66.

STEP 4

List the factor pairs of 27-27 and find the pair that sums to 66:
Possible pairs: (1,27),(1,27),(3,9),(3,9)(-1, 27), (1, -27), (-3, 9), (3, -9).
The pair (3,9)(-3, 9) satisfies the condition 3+9=6-3 + 9 = 6.

STEP 5

Write the factorization using the identified values:
x2+6x27=(x(3))(x+9) x^2 + 6x - 27 = (x - (-3))(x + 9)
Simplify the expression:
(x+3)(x+9) (x + 3)(x + 9)

STEP 6

Verify the factorization by expanding:
(x+3)(x+9)=x2+9x+3x+27=x2+12x+27 (x + 3)(x + 9) = x^2 + 9x + 3x + 27 = x^2 + 12x + 27
The factorization is incorrect. Re-evaluate the factor pairs.

STEP 7

Re-evaluate the factor pairs of 27-27 and find the correct pair:
The correct pair should be (3,9)(3, -9) because:
3(9)=27 3 \cdot (-9) = -27 3+(9)=6 3 + (-9) = -6
The signs were incorrect in the previous step.

STEP 8

Write the correct factorization:
x2+6x27=(x3)(x+9) x^2 + 6x - 27 = (x - 3)(x + 9)
Verify by expanding:
(x3)(x+9)=x2+9x3x27=x2+6x27 (x - 3)(x + 9) = x^2 + 9x - 3x - 27 = x^2 + 6x - 27
The factorization is now correct.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord