Factorise completely the following quadratic expressions:

a. x^2 + 2x - 3

b. 6 - 5x - x^2

c. 3x^2 - 17x + 10

d. p^2 - 1

The quadratic expressions can be factorized as follows: x 2 + 2 x − 3 = ( x + 3 ) ( x − 1 ) , 6 − 5 x − x 2 = − 1 ( x + 6 ) ( x − 1 ) , 3 x 2 − 17 x + 10 = ( 3 x − 2 ) ( x − 5 ) , and p 2 − 1 = ( p − 1 ) ( p + 1 ) . Each factorization approaches different types of problems such as grouping, difference of squares, and finding products that meet sum and product criteria. Understanding these techniques is essential in solving quadratic expressions.
;

Answered by Anonymous | 2025-07-04

To factorise quadratic expressions completely, we need to express each quadratic as a product of two binomials. Let's walk through the factorisation of these quadratics step-by-step:
a. x 2 + 2 x − 3
To factorise x 2 + 2 x − 3 , we look for two numbers that multiply to − 3 (the constant term) and add up to 2 (the coefficient of x ). These numbers are 3 and − 1 .
Thus, we can write: x 2 + 2 x − 3 = ( x + 3 ) ( x − 1 )
b. 6 − 5 x − x 2
First, let's rewrite the expression in a standard quadratic form: − x 2 − 5 x + 6 Alternatively, you can rearrange terms to make it recognizable, which is: − x 2 − 5 x + 6 = − ( x 2 + 5 x − 6 ) Now, factor x 2 + 5 x − 6 as before by finding two numbers that multiply to − 6 and add up to 5 . These numbers are 6 and − 1 .
Thus, the factorised form is: − ( x + 6 ) ( x − 1 )
c. 3 x 2 − 17 x + 10
We need two numbers that multiply to 3 × 10 = 30 and add up to − 17 . Those numbers are − 15 and − 2 .
We rewrite the middle term using these numbers: 3 x 2 − 15 x − 2 x + 10 Now, group the terms: 3 x ( x − 5 ) − 2 ( x − 5 ) Factor by grouping gives us: ( 3 x − 2 ) ( x − 5 )
d. p 2 − 1
This is a difference of squares, which can be factorised using the formula a 2 − b 2 = ( a − b ) ( a + b ) . Here, a = p and b = 1 , so: p 2 − 1 = ( p − 1 ) ( p + 1 )
I hope this helps you understand how to factorise these quadratic expressions!

Answered by OliviaMariThompson | 2025-07-06