Core Pure -as Year 1- Unit Test 5 Algebra And Functions Apr 2026

She felt a small smile. But the test wasn't done.

She wrote: No solution (the expression is always ≥ 0). A trick question. But she didn't fall for it.

And for the first time, she felt like a real mathematician.

One down.

hit her like a cold splash of water. Given that ( f(x) = 2x^3 + 3x^2 - 8x + 3 ), show that ( (x-1) ) is a factor, and hence fully factorise ( f(x) ). Elena took a breath. Polynomials. I can do this. She scribbled the substitution: ( f(1) = 2 + 3 - 8 + 3 = 0 ). Yes. Then came the algebraic long division, the careful subtraction of terms, the descent into the quadratic. ( (x-1)(2x^2 + 5x - 3) ). Then the final break: ( (x-1)(2x-1)(x+3) ).

Never. A square of a real number is always ( \geq 0 ). The only time it equals zero is at the roots. So no real ( x ) satisfies ( p(x) < 0 ).

She flipped back. Question 6 (not mentioned yet) was a proof by contradiction involving a rational root of a cubic. She had left it till last. Prove that ( \sqrt{3} ) is irrational. She wrote: Assume ( \sqrt{3} = \frac{a}{b} ) in lowest terms. Then ( 3b^2 = a^2 ). So 3 divides ( a^2 ), so 3 divides ( a ). Let ( a = 3k ). Then ( 3b^2 = 9k^2 ) → ( b^2 = 3k^2 ). So 3 divides ( b^2 ), so 3 divides ( b ). Contradiction — ( a ) and ( b ) have a common factor 3, not lowest terms. Hence ( \sqrt{3} ) is irrational. core pure -as year 1- unit test 5 algebra and functions

was the killer. The one that separated the A from the B. The function ( p(x) = x^4 - 8x^2 + 16 ). Find all real roots. Hence solve the inequality ( p(x) < 0 ). She factorised: let ( u = x^2 ). Then ( u^2 - 8u + 16 = (u-4)^2 ). So ( p(x) = (x^2 - 4)^2 = (x-2)^2 (x+2)^2 ).

The answer formed: ( \frac{1}{x-1} - \frac{1}{x+2} + \frac{5}{x-3} ). Clean. Elegant.

As she walked out, she thought: That wasn't a test. That was a rite of passage. She felt a small smile

Unit Test 5 wasn't just about algebra. It was about precision. About checking every assumption. About remembering that a square can never be negative.

But the domain of ( h \circ k ) is ( { x \in \text{dom}(k) \mid k(x) \in \text{dom}(h) } ). ( x \geq 0 ) and ( x^2 - 1 \geq -4 ) — which is always true. So the domain is simply ( x \geq 0 ).

She turned the page.