The factorisation of – x2 + 5x – 6 yields (a) – (x – 2)(3 – x) (b) – (2 – x) (3 – x) (c) (x – 2)(x – 3) (d) (2 + x) (3 – x)
The Correct Answer and Explanation is:
The correct answer is: (b) – (2 – x)(3 – x)
Explanation:
Start by analyzing the given quadratic expression:−x2+5x−6- x^2 + 5x – 6−x2+5x−6
To factor this expression, consider removing the negative sign for easier handling:−(x2−5x+6)- (x^2 – 5x + 6)−(x2−5x+6)
Now focus on factoring the expression inside the parentheses:x2−5x+6x^2 – 5x + 6×2−5x+6
This quadratic has a standard form ax2+bx+cax^2 + bx + cax2+bx+c, where:
- a=1a = 1a=1
- b=−5b = -5b=−5
- c=6c = 6c=6
Look for two numbers that multiply to 6 and add to -5. These numbers are:
- -2 and -3
Therefore, the expression factors as:x2−5x+6=(x−2)(x−3)x^2 – 5x + 6 = (x – 2)(x – 3)x2−5x+6=(x−2)(x−3)
So the original expression becomes:−(x−2)(x−3)- (x – 2)(x – 3)−(x−2)(x−3)
This form can be manipulated into different but equivalent forms. Applying the property of signs:−(x−2)(x−3)=−(2−x)(3−x)- (x – 2)(x – 3) = – (2 – x)(3 – x)−(x−2)(x−3)=−(2−x)(3−x)
This works because:x−2=−(2−x),x−3=−(3−x)x – 2 = – (2 – x), \quad x – 3 = – (3 – x)x−2=−(2−x),x−3=−(3−x)
So:(x−2)(x−3)=[−(2−x)][−(3−x)]=(2−x)(3−x)(x – 2)(x – 3) = [ – (2 – x) ] [ – (3 – x) ] = (2 – x)(3 – x)(x−2)(x−3)=[−(2−x)][−(3−x)]=(2−x)(3−x)
However, due to the negative outside the brackets, it results in:−(x−2)(x−3)=−(2−x)(3−x)- (x – 2)(x – 3) = – (2 – x)(3 – x)−(x−2)(x−3)=−(2−x)(3−x)
Thus, the equivalent and correct factorization is:
– (2 – x)(3 – x)
This matches option (b).
