Question
If $p(y)=2 y^3-6 y^2-5 y+7$, then find $p(2)$.
✓
Answer
$p(y)=2 y^3-6 y^2-5 y+7$
Put $y=2$ in the given polynomial.
$\therefore p(2)=2(2)^3-6(2)^2-5(2)+7$
$=2 \times 8-6 \times 4-10+7$
$=16-24-10+7$
$\therefore P(2)=-11$
Put $y=2$ in the given polynomial.
$\therefore p(2)=2(2)^3-6(2)^2-5(2)+7$
$=2 \times 8-6 \times 4-10+7$
$=16-24-10+7$
$\therefore P(2)=-11$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Define the following terms:
Concurrent lines.
Concurrent lines.
Find the reduced form of the ratio of the first quantity to second quantity : 5 litres, 2500 ml
When one pair of interior angles on one side of the transversal is supplementary, then the other pair of interior angles is also supplementary.
Write the following in the expanded form:
$(2+x-2 y)^2$
→$(2+x-2 y)^2$
The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
In the right angled ∆XYZ, ∠XYZ = 90° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios.
i. sin x
ii. tan z
iii. cos x
iv. tan x.
Factorise:
$25 x^2+4 y^2+9 z^2-20 x y-12 y z+30 x z$
→$25 x^2+4 y^2+9 z^2-20 x y-12 y z+30 x z$
Factorise:$\frac{3}{2}\text{x}^2+16\text{x}+10$
→Factorise:
$81 x^4-y^4$
→$81 x^4-y^4$
Prove the Theorem : In a right angled triangle, the length of the median of the hypotenuse is half the length of the hypotenuse.